NCERT Solutions for Class 9 Mathematics Chapter 7- Triangles

Q.1

$\begin{array}{l}\mathrm{In}\mathrm{quadrilateral}\mathrm{ACBD},\mathrm{AC}=\mathrm{AD}\mathrm{and}\mathrm{AB}\mathrm{bisects}\mathrm{}\angle \mathrm{A}\\ (\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{Show}\mathrm{that}\mathrm{\Delta ABC}\cong \mathrm{\Delta ABD}.\mathrm{}\mathrm{What}\mathrm{can}\\ \mathrm{you}\mathrm{say}\mathrm{about}\mathrm{BC}\mathrm{and}\mathrm{BD}.\end{array}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{ABCD is a quadrilateral, AC}=\text{BD, AB bisects}\angle \text{A.}\\ \text{To prove:\hspace{0.17em}}\mathrm{\Delta }\text{ABC}\cong \mathrm{\Delta }\text{ABD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:In}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{ABD}\\ \text{AC}=\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \angle \mathrm{BAC}=\angle \mathrm{ABD}\left[\text{Given}\right]\\ \mathrm{AB}=\mathrm{AB}\left[\text{Common}\right]\\ \therefore \mathrm{\Delta }\text{ABC}\cong \mathrm{\Delta }\text{ABD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{SAS}\right]\\ \mathrm{BC}=\mathrm{BD}\left[\mathrm{By}\text{CPCT}\right]\end{array}$

Q.2

$\begin{array}{l}\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{quadrilateral}\mathrm{in}\mathrm{which}\mathrm{AD}=\mathrm{BC}\mathrm{and}\angle \mathrm{\hspace{0.17em}}\mathrm{DAB}=\angle \mathrm{\hspace{0.17em}}\mathrm{CBA}\\ (\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{Prove}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{\Delta ABD}\cong \mathrm{\Delta BAC}\\ \left(\mathrm{ii}\right)\mathrm{BD}=\mathrm{AC}\\ \left(\mathrm{iii}\right)\angle \mathrm{ABD}=\angle \mathrm{\hspace{0.17em}}\mathrm{BAC}.\end{array}$

Ans.

$\begin{array}{l}\text{Given}:\text{ABCD is a quadrilateral, AD}=\text{BC,}\angle D\text{AB}=\angle C\text{BA}\text{.}\\ \text{To prove:}\\ \left(i\right)\Delta \text{ABD}\cong \Delta B\text{AC}\\ \left(ii\right)BD=AC\\ \left(iii\right)\angle \text{ABD}=\angle \text{BAC}\\ \text{Proof:}\left(i\right)\text{In}\Delta \text{ABD and}\Delta B\text{AC}\\ \text{AB}=B\text{A}\left[\text{Common}\right]\\ \angle BAD=\angle ABC\left[\text{Given}\right]\\ AD=BC\left[\text{Given}\right]\\ \therefore \Delta \text{ABD}\cong \Delta \text{BAC}\left[By\text{SAS}\right]\\ \left(ii\right)BD=AC\left[By\text{CPCT}\right]\\ \left(iii\right)\angle \text{ABD}=\angle \text{BAC}\left[By\text{CPCT}\right]\\ \end{array}$

Q.3 AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Ans.

$\begin{array}{l}\text{Given}:\text{AD and BC are equal perpendiculars to a line segment AB.}\\ \text{To Prove: CD bisects AB i.e., CO}=\text{OD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\mathrm{In}\text{}\mathrm{\Delta BOC}\cong \mathrm{\Delta AOD}\\ \angle \mathrm{CBO}=\angle \mathrm{DAO}\left[\mathrm{Given}\right]\\ \angle \mathrm{BOC}=\angle \mathrm{AOD}\left[\mathrm{Vertical}\text{opposite angles}\right]\\ \mathrm{BC}=\mathrm{AD}\left[\mathrm{Given}\right]\\ \therefore \mathrm{\Delta BOC}\cong \mathrm{\Delta AOD}\left[\mathrm{By}\text{AAS}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BO}=\text{OA\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \Rightarrow \mathrm{CD}\text{bisects line segment AB.}\end{array}$

Q.4

$\begin{array}{l}\mathrm{l}\mathrm{and}\mathrm{m}\mathrm{are}\mathrm{two}\mathrm{parallell}\mathrm{ines}\mathrm{intersected}\mathrm{by}\mathrm{another}\mathrm{pair}\mathrm{of}\\ \mathrm{parallel}\mathrm{lines}\mathrm{p}\mathrm{and}\mathrm{q}(\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{Show}\mathrm{that}\mathrm{\Delta }\mathrm{ABC}\cong \mathrm{\Delta CDA}.\end{array}$

Ans.

$\begin{array}{l}\text{Given}:\text{line}\mathrm{l}\text{and m are parallel and}\stackrel{\rightharpoonup }{\mathrm{p}}\parallel \stackrel{\rightharpoonup }{\mathrm{q}}.\\ \text{To prove:\hspace{0.17em}}\mathrm{\Delta }\mathrm{ABC}\cong \mathrm{\Delta CDA}\\ \text{Proof: In\hspace{0.17em}}\mathrm{\Delta }\mathrm{ABCand}\mathrm{\Delta CDA}\\ \angle \mathrm{BAC}=\angle \mathrm{DCA}\left[\mathrm{Alternate}\text{\hspace{0.17em}interior angles}\right]\\ \mathrm{AC}=\mathrm{CA}\left[\text{Common}\right]\\ \angle \mathrm{BCA}=\angle \mathrm{DAC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Alternate\hspace{0.17em}interior angles]}\\ \mathrm{\Delta }\mathrm{ABC}\cong \mathrm{\Delta CDA}\left[\mathrm{By}\text{ ASA}\right]\end{array}$

Q.5

$\begin{array}{l}\text{Line \hspace{0.17em}l \hspace{0.17em}is the bisector of an angle \hspace{0.17em}∠A and ∠B is any point on \hspace{0.17em}l.}\\ \text{BP and BQ are perpendiculars \hspace{0.17em}from B to the arms of\hspace{0.17em}∠A}\\ (\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{}\mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\mathrm{\Delta APB}\cong \mathrm{\Delta AQB}\mathrm{}\\ \left(\mathrm{ii}\right)\mathrm{BP}=\mathrm{BQ}\mathrm{or}\mathrm{B}\mathrm{is}\mathrm{equidistant}\mathrm{from}\mathrm{the}\mathrm{arms}\hspace{0.17em}\mathrm{of}\angle \mathrm{A}.\end{array}$

Ans.

$\begin{array}{l}\text{Given}:\text{Line}\mathrm{l}\text{is bisector of}\angle \text{A, BP and BQ are perpendiculars to}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}the arms of angle.}\\ \mathrm{To}\text{prove:}\left(\mathrm{i}\right)\mathrm{\Delta }\mathrm{APB}\cong \mathrm{\Delta AQB}\\ \left(\mathrm{ii}\right)\mathrm{BP}=\mathrm{BQ}\mathrm{or}\mathrm{B}\text{is equidistant from the arms \hspace{0.17em}of∠A.}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In\hspace{0.17em}}\mathrm{\Delta }\mathrm{ABQ}\mathrm{and}\mathrm{\Delta APB}\\ \angle \mathrm{BAQ}=\angle \mathrm{BAP}\left[\mathrm{Given}\right]\\ \mathrm{AB}=\mathrm{AB}\left[\mathrm{Common}\right]\\ \angle \mathrm{AQB}=\angle \mathrm{APB}\left[\mathrm{each}\text{90°}\right]\\ \mathrm{\Delta }\mathrm{ABQ}\cong \mathrm{\Delta APB}\left[\mathrm{By}\text{ AAS}\right]\\ \left(\mathrm{ii}\right)\mathrm{BQ}=\mathrm{BP}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \Rightarrow \mathrm{BP}=\mathrm{BQ}\end{array}$

Q.6 In figure below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{In}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{ADE, AC}=\text{AE},\text{AB}=\text{AD and}\angle \text{BAD}=\angle \text{EAC}\\ \text{To prove:BC=DE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof: In\hspace{0.17em}}\mathrm{\Delta }\mathrm{ABC}\mathrm{and}\mathrm{\Delta ADE}\\ \mathrm{AB}=\mathrm{AD}\left[\text{Given}\right]\\ \angle \mathrm{BAC}=\angle \mathrm{DAE}[\angle \mathrm{BAD}+\angle \mathrm{DAC}=\angle \mathrm{DAC}+\angle \mathrm{EAC}]\\ \mathrm{AC}=\mathrm{AE}\left[\mathrm{Given}\right]\\ \mathrm{\Delta }\mathrm{ABC}\cong \mathrm{\Delta ADE}\left[\mathrm{By}\text{ SAS}\right]\\ \mathrm{BC}=\mathrm{DE}\left[\mathrm{By}\text{C.P.C.T.}\right]\end{array}$

Q.7

$\begin{array}{l}\mathrm{AB}\mathrm{is}\mathrm{a}\mathrm{line}\mathrm{segment}\mathrm{and}\mathrm{P}\mathrm{is}\mathrm{its}\mathrm{mid}–\mathrm{point}.\mathrm{D}\mathrm{and}\mathrm{E}\mathrm{are}\\ \mathrm{points}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{AB}\mathrm{such}\mathrm{that}\angle \mathrm{BAD}=\angle \mathrm{ABE}\\ \mathrm{and}\angle \mathrm{EPA}=\angle \mathrm{DPB}(\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{\Delta DAP}\cong \mathrm{\Delta EBP}\\ \left(\mathrm{ii}\right)\mathrm{AD}=\mathrm{BE}\end{array}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{P}\text{is mid-point of AB,\hspace{0.17em}}\angle \text{BAD}=\angle \mathrm{ABE}\text{ and}\angle \mathrm{EPA}=\angle \mathrm{DPB}\\ \text{To prove:}\left(\mathrm{i}\right)\mathrm{\Delta DAP}\cong \mathrm{\Delta EBP}\\ \left(\mathrm{ii}\right)\mathrm{AD}=\mathrm{BE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\left(\mathrm{i}\right)\text{In\hspace{0.17em}}\mathrm{\Delta }\mathrm{DAPand}\mathrm{\Delta EBP}\\ \angle \mathrm{DPA}=\angle \mathrm{EPB}[\angle \mathrm{APE}+\angle \mathrm{EPD}=\angle \mathrm{BPD}+\angle \mathrm{DPE}]\\ \mathrm{A}\mathrm{P}=\mathrm{PB}\left[\mathrm{Given}\right]\\ \angle \mathrm{DAP}=\angle \mathrm{EBP}\left[\mathrm{Given}\right]\\ \mathrm{\Delta }\mathrm{DAP}\cong \mathrm{\Delta EPB}\left[\mathrm{By}\text{ SAS}\right]\\ \left(\mathrm{ii}\right)\mathrm{AD}=\mathrm{BE}\left[\mathrm{By}\text{C.P.C.T.}\right]\end{array}$

Q.8

$\begin{array}{l}\mathrm{In}\mathrm{right}\mathrm{triangle}\mathrm{ABC},\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{C},\mathrm{M}\mathrm{is}\mathrm{the}\mathrm{mid}–\mathrm{point}\\ \mathrm{of}\mathrm{hypotenuse}\mathrm{AB}.\mathrm{C}\mathrm{is}\mathrm{joined}\mathrm{to}\mathrm{M}\mathrm{and}\mathrm{produced}\mathrm{to}\mathrm{a}\mathrm{point}\\ \mathrm{D}\mathrm{such}\mathrm{that}\mathrm{DM}=\mathrm{CM}.\mathrm{Point}\mathrm{D}\mathrm{is}\mathrm{joined}\mathrm{to}\mathrm{point}\mathrm{B}\\ (\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\mathrm{\Delta AMC}\cong \mathrm{\Delta BMD}\\ \left(\mathrm{ii}\right)\mathrm{\Delta DBC}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}.\\ \left(\mathrm{iii}\right)\mathrm{\Delta DBC}\cong \mathrm{\Delta ACB}\\ \left(\mathrm{iv}\right)\mathrm{CM}=\frac{1}{2}\mathrm{AB}\end{array}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{In right}\mathrm{\Delta }\text{ABC,}\angle \text{C}=90\mathrm{°}\text{and M is mid-point on}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}hypotenuse AB,\hspace{0.17em}CM}=\text{DM.}\\ \text{To Prove:}\left(\mathrm{i}\right)\mathrm{\Delta AMC}\cong \mathrm{\Delta BMD}\\ \left(\mathrm{ii}\right)\angle \mathrm{DBC}\text{is a right angle.}\\ \left(\mathrm{iii}\right)\mathrm{\Delta DBC}\cong \mathrm{\Delta ACB}\\ \left(\mathrm{iv}\right)\mathrm{CM}=\frac{1}{2}\mathrm{AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\left(\mathrm{i}\right)\text{In}\mathrm{\Delta AMC}\mathrm{and}\mathrm{\Delta BMD}\\ \mathrm{AM}=\mathrm{BM}\left[\mathrm{Given}\right]\\ \angle \mathrm{AMC}=\angle \mathrm{BMD}\left[\text{Vertica}\mathrm{l}\text{opposite angles}\right]\\ \mathrm{CM}=\mathrm{DM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Given]}\end{array}$ $\begin{array}{l}\therefore \mathrm{\Delta AMC}\cong \mathrm{\Delta BMD}[\mathrm{S}.\mathrm{A}.\mathrm{S}.]\\ \mathrm{AC}=\mathrm{DB}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \left(\mathrm{ii}\right)\angle \mathrm{ACM}=\angle \mathrm{BDM}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \Rightarrow \angle \mathrm{ACD}=\angle \mathrm{BDC}\\ \Rightarrow \mathrm{AC}\parallel \mathrm{DB}\left[\mathrm{Alternate}\text{angles are equal.}\right]\\ \angle \mathrm{ACB}+\angle \mathrm{DBC}=180\mathrm{°}\left[\text{Co-interior angles}\right]\\ 90\mathrm{°}+\angle \mathrm{DBC}=180\mathrm{°}\\ \angle \mathrm{DBC}=180\mathrm{°}-90\mathrm{°}\\ =90\mathrm{°}\\ \mathrm{So},\angle \mathrm{DBC}\text{is a right angle.}\\ \left(\mathrm{iii}\right)\mathrm{In}\mathrm{\Delta }\text{DCB and}\mathrm{\Delta ACB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}DB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Proved}\right]\\ \angle \mathrm{ACB}=\angle \mathrm{DBC}\left[\mathrm{Each}\text{90°}\right]\\ \mathrm{BC}=\mathrm{BC}\left[\mathrm{Common}\right]\\ \therefore \mathrm{\Delta }\text{DCB}\cong \mathrm{\Delta ACB}[\mathrm{S}.\mathrm{A}.\mathrm{S}.]\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}DC}=\text{AB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ 2\mathrm{CM}=\mathrm{AB}[\mathrm{CM}=\mathrm{DM}]\\ \mathrm{CM}=\frac{1}{2}\mathrm{AB}\end{array}$

Q.9 In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC
(ii) AO bisects ∠A.

Ans.

$\begin{array}{l}\text{Given: In}\mathrm{\Delta }\text{ABC, AB}=\text{AC and bisectors of}\angle \text{B and}\angle \text{C intersect}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}at O.}\\ \text{To Prove:}\left(\text{i}\right)\text{OB}=\text{OC}\left(\text{ii}\right)\text{AO bisects}\angle \text{A}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In}\mathrm{\Delta }\text{ABC, AB}=\text{AC}\\ \text{So, \hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}=\angle \mathrm{ABC}\left[\begin{array}{l}\text{Opposite angles of equal}\\ \text{sides are equal.}\end{array}\right]\\ \Rightarrow \frac{1}{2}\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{ABC}\\ \Rightarrow \angle \mathrm{OCB}=\angle \mathrm{OBC}\left[\begin{array}{l}\text{Given},\text{as OB and OC are angle}\\ \text{bisectors.}\end{array}\right]\\ \Rightarrow \mathrm{OB}=\mathrm{OC}\left[\begin{array}{l}\text{Opposite angles of equal}\\ \text{sides are equal.}\end{array}\right]\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\text{In}\mathrm{\Delta O}\text{AB and}\mathrm{\Delta O}\text{AC,}\\ \mathrm{OA}=\mathrm{OA}\left[\mathrm{Common}\right]\\ \mathrm{AB}=\mathrm{AC}\left[\mathrm{Given}\right]\\ \mathrm{OB}=\mathrm{OC}\left[\text{Proved above}\right]\\ \therefore \mathrm{\Delta O}\text{AB}\cong \mathrm{\Delta O}\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \mathrm{Hence},\angle \mathrm{OAB}=\angle \mathrm{OAC}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{Thus},\text{OA bisects}\angle \text{A.}\end{array}$

Q.10 In Δ ABC, AD is the perpendicular bisector of BC (see figure below). Show that Δ ABC is an isosceles triangle in which AB = AC.

Ans.

$\begin{array}{l}\text{Given}:\text{In}\mathrm{\Delta }\text{ABC, BD}=\text{DC and AD}\perp \text{BC.}\\ \text{To Prove: AB}=\text{AC}\\ \text{Proof: In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{ACD}\\ \mathrm{AD}=\mathrm{AD}\left[\text{Common}\right]\\ \angle \mathrm{ADB}=\angle \mathrm{ADC}\left[\text{Given}\right]\\ \mathrm{B}\mathrm{D}=\mathrm{DC}[\text{Given]}\\ \therefore \mathrm{\Delta }\text{ABD}\cong \mathrm{\Delta }\text{ACD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \therefore \mathrm{AB}=\mathrm{AC}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Therefore},\mathrm{\Delta }\text{ABC is isosceles triangle.}\end{array}$

Q.11 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure below). Show that these altitudes are equal.

Ans.

$\begin{array}{l}\text{Given:In\hspace{0.17em}}\mathrm{\Delta ABC},\mathrm{AC}=\mathrm{AB}\text{, BE}\perp \text{AC and CF}\perp \text{AB.}\\ \text{To prove: BE}=\text{CF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\mathrm{In}\mathrm{\Delta ABE}\text{and\hspace{0.17em}}\mathrm{\Delta ACF}\\ \angle \mathrm{BAE}=\angle \mathrm{CAF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Common angle]}\\ \angle \mathrm{BEA}=\angle \mathrm{CFA}\left[\mathrm{Each}\text{90°}\right]\\ \mathrm{AB}=\mathrm{AC}[\text{Given]}\\ \therefore \mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}\left[\mathrm{By}\text{A.A.S.}\right]\\ \therefore \mathrm{BE}=\mathrm{CF}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{in}\mathrm{which}\mathrm{altitudes}\mathrm{BE}\mathrm{and}\mathrm{CF}\mathrm{to}\mathrm{sides}\mathrm{AC}\\ \mathrm{and}\mathrm{AB}\mathrm{are}\mathrm{equal}(\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}\\ \left(\mathrm{ii}\right)\mathrm{AB}=\mathrm{AC},\mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{\hspace{0.17em}}\mathrm{Given}:\mathrm{In}\mathrm{\hspace{0.17em}}\mathrm{\Delta ABC},\mathrm{\hspace{0.17em}}\mathrm{BE}=\mathrm{CF},\mathrm{BE}\perp \mathrm{AC}\mathrm{and}\mathrm{CF}\perp \mathrm{AB}\mathrm{.}\\ \mathrm{To}\mathrm{prove}:\left(\mathrm{i}\right)\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{ii}\right)\mathrm{AB}=\mathrm{AC},\mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Proof}:\left(\mathrm{i}\right)\mathrm{In}\mathrm{\hspace{0.17em}}\mathrm{\Delta ABE}\mathrm{and}\mathrm{\hspace{0.17em}}\mathrm{\Delta ACF}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\angle \mathrm{BAE}=\angle \mathrm{CAF}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}[\mathrm{Common}\mathrm{angle}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\angle \mathrm{BEA}=\angle \mathrm{CFA}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}[\mathrm{Each}90°]\\ \mathrm{BE}=\mathrm{CF}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\Delta ABE}\cong \mathrm{\hspace{0.17em}}\mathrm{\Delta ACF}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\mathrm{By}\mathrm{A}.\mathrm{A}.\mathrm{S}\mathrm{.}]\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\mathrm{.}]\\ \mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Hence}\mathrm{proved}\mathrm{.}\end{array}$

Q.13 ABC and DBC are two isosceles triangles on the same base BC (see figure below). Show that ∠ABD = ∠ACD.

Ans.

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\mathrm{\Delta ABC},\mathrm{AB}=\mathrm{AC}\text{and}\mathrm{in}\mathrm{\Delta DBC},\mathrm{DB}=\mathrm{DC}\text{.}\\ \text{To prove:}\angle \mathrm{ABD}=\angle \mathrm{ACD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\mathrm{In}\mathrm{\Delta ABC}\\ \mathrm{AB}=\mathrm{AC}\left[\mathrm{Given}\right]\\ \therefore \angle \mathrm{ACB}=\angle \mathrm{ABC}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{In}\mathrm{\Delta DBC}\\ \mathrm{DB}=\mathrm{DC}\left[\mathrm{Given}\right]\\ \therefore \angle \mathrm{DCB}=\angle \mathrm{DBC}...\left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \angle \mathrm{ACB}+\angle \mathrm{DCB}=\angle \mathrm{ABC}+\angle \mathrm{DBC}\\ \therefore \angle \mathrm{ACD}=\angle \mathrm{ABD}\\ \Rightarrow \angle \mathrm{ABD}=\angle \mathrm{ACD}\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.14 ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure below). Show that ∠BCD is a right angle.

Ans.

$\begin{array}{l}\text{Given:In}\mathrm{\Delta }\text{ABC, AB}=\text{AC, side BA is produced to D such that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}=\text{AB.}\\ \text{To prove:}\angle \text{BCD is a right angle}.\\ \text{Proof:In}\mathrm{\Delta }\text{ABC,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}[\text{Given]}\\ \angle \text{ACB}=\angle \text{ABC \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite angles of equal sides are equal.}\right]\\ =\mathrm{x}\left(\mathrm{let}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta }\text{ACD,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}=\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Given]}\\ \angle \text{ADC}=\angle \text{ACD \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite angles of equal sides are equal.}\right]\\ =\mathrm{y}\left(\mathrm{let}\right)\\ \text{Now},\text{\hspace{0.17em}In}\mathrm{\Delta }\text{BCD}\\ \angle \text{BCD}+\angle \mathrm{C}\text{BD}+\angle \text{BDC}=180\mathrm{°}\\ (\mathrm{x}+\mathrm{y})+\mathrm{x}+\mathrm{y}=180\mathrm{°}\\ 2(\mathrm{x}+\mathrm{y})=180\mathrm{°}\\ (\mathrm{x}+\mathrm{y})=\frac{180\mathrm{°}}{2}\\ \angle \text{BCD}=90\mathrm{°}\\ \Rightarrow ∆\text{BCD is right triangle.}\end{array}$

Q.15 ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}Given :In}\mathrm{\Delta }\text{ABC,}\angle \text{A}=\text{90° and AB}=\text{AC.}\\ \text{To find :\hspace{0.17em}}\angle \text{\hspace{0.17em}B and}\angle \text{\hspace{0.17em}C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sol : In}\mathrm{\Delta }\text{ABC,}\\ \mathrm{AB}=\mathrm{AC}[\text{Given]}\\ \mathrm{So},\angle \mathrm{ACB}=\angle \mathrm{ABC}=\mathrm{x}\left(\mathrm{let}\right)\\ \text{Since},\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ 90\mathrm{°}+\mathrm{x}+\mathrm{x}=180\mathrm{°}\\ 2\mathrm{x}=180\mathrm{°}-90\mathrm{°}\\ \mathrm{x}=\frac{90\mathrm{°}}{2}=45\mathrm{°}\\ \mathrm{So},\angle \mathrm{B}=45\mathrm{°}\text{and}\angle \mathrm{C}=45\mathrm{°}.\end{array}$

Q.16 Show that the angles of an equilateral triangle are 60° each.

Ans.

$\begin{array}{l}\mathrm{Given}:\mathrm{\Delta ABC}\text{is an equilateral triangle.}\\ \text{To prove :}\angle \text{A}=\angle \mathrm{B}=\angle \text{C}=60\mathrm{°}\\ \text{Proof}:\text{In}\mathrm{\Delta ABC},\\ \mathrm{AB}=\mathrm{AC}\left[\mathrm{Given}\right]\end{array}$

$\begin{array}{l}\angle \mathrm{C}=\angle \mathrm{B}...\left(\mathrm{i}\right)\left[\begin{array}{l}\text{Angles opposite to equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{AB}=\mathrm{BC}\left[\mathrm{Given}\right]\\ \angle \mathrm{C}=\angle \mathrm{A}...\left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{Angles}\text{opposite to equal sides}\\ \text{are equal.}\end{array}\right]\\ \text{By angle sum property in a triangle,}\\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ \angle \mathrm{C}+\angle \mathrm{C}+\angle \mathrm{C}=180\mathrm{°}\left[\text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right)\right]\\ 3\angle \mathrm{C}=180\mathrm{°}\\ \angle \mathrm{C}=\frac{180\mathrm{°}}{3}=60\mathrm{°}\\ \mathrm{So},\angle \mathrm{A}=60\mathrm{°}\text{and}\angle \mathrm{B}=60\mathrm{°}\\ \text{Therefore},\text{each angle of an equilateral triangle is 60° each.}\end{array}$

Q.17

$\begin{array}{l}\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta DBC}\mathrm{are}\mathrm{two}\mathrm{isosceles}\mathrm{triangle}\mathrm{son}\mathrm{the}\mathrm{same}\\ \mathrm{base}\mathrm{BC}\mathrm{and}\mathrm{vertices}\mathrm{A}\mathrm{and}\mathrm{D}\mathrm{are}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{BC}\\ (\mathrm{see}\mathrm{the}\mathrm{figure}\mathrm{below}).\mathrm{If}\mathrm{AD}\mathrm{is}\mathrm{extended}\mathrm{to}\mathrm{intersect}\mathrm{BC}\mathrm{at}\mathrm{P},\mathrm{show}\\ \text{that}\end{array}$ $\begin{array}{l}\left(\mathrm{i}\right)\mathrm{\Delta ABD}\cong \mathrm{\Delta ACD}\\ \left(\mathrm{ii}\right)\mathrm{\Delta ABP}\cong \mathrm{\Delta ACP}\\ \left(\mathrm{iii}\right)\mathrm{AP}\mathrm{bisects}\angle \mathrm{A}\mathrm{as}\mathrm{well}\mathrm{as}\angle \mathrm{D}.\\ \left(\mathrm{iv}\right)\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{BC}.\end{array}$

Ans.

$\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{Given}:\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta DBC}\mathrm{are}\mathrm{two}\mathrm{isosceles}\mathrm{triangles}\mathrm{on}\mathrm{the}\mathrm{same}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{base}\mathrm{BC}\mathrm{}\mathrm{and}\mathrm{vertices}\mathrm{A}\mathrm{and}\mathrm{D}\mathrm{are}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{BC}.\\ \mathrm{To}\mathrm{prove}:\mathrm{\hspace{0.17em}}\left(\mathrm{i}\right)\mathrm{\Delta ABD}\cong \mathrm{\Delta ACD}\\ \mathrm{\hspace{0.17em}}\left(\mathrm{ii}\right)\mathrm{\Delta ABP}\cong \mathrm{\Delta ACP}\\ \mathrm{\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{AP}\mathrm{bisects}\angle \mathrm{A}\mathrm{as}\mathrm{well}\mathrm{as}\angle \mathrm{D}.\\ \\ \mathrm{\hspace{0.17em}}\left(\mathrm{iv}\right)\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{BC}.\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Proof}:\left(\mathrm{i}\right)\mathrm{In}\mathrm{}\mathrm{\Delta ABD}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\mathrm{\Delta ACD}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{AB}=\mathrm{AC}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{DB}=\mathrm{DC}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{AD}=\mathrm{AD}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \therefore \mathrm{\Delta ABD}\mathrm{\hspace{0.17em}}\cong \mathrm{\Delta ACD}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}[\mathrm{By}\mathrm{S}.\mathrm{S}.\mathrm{S}\mathrm{.}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\angle \mathrm{BAD}=\angle \mathrm{CAD}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}[\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\mathrm{.}]\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\mathrm{In}\mathrm{\Delta ABP}\mathrm{and\Delta ACP}\\ \mathrm{AB}=\mathrm{AC}\left[\mathrm{Given}\right]\\ \angle \mathrm{BAP}=\angle \mathrm{CAP}\left[\text{Proved above}\right]\\ \mathrm{AP}=\mathrm{AP}\left[\mathrm{Common}\right]\\ \therefore \mathrm{\Delta ABP}\cong \mathrm{\Delta ACP}\left[\mathrm{By}\text{S.A.S.}\right]\\ \angle \mathrm{BAP}=\angle \mathrm{CAP}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{BP}=\mathrm{CP}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \left(\mathrm{iii}\right)\because \angle \mathrm{BAP}=\angle \mathrm{CAP}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \Rightarrow \mathrm{AP}\text{bisects}\angle \mathrm{A}.\\ \mathrm{In}\mathrm{\Delta BDP}\mathrm{and\Delta }\mathrm{CDP}\\ \mathrm{BD}=\mathrm{CD}\left[\mathrm{Given}\right]\\ \mathrm{DP}=\mathrm{DP}\left[\mathrm{Common}\right]\\ \mathrm{BP}=\mathrm{CP}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \therefore \mathrm{\Delta BDP}\cong \mathrm{\Delta }\mathrm{CDP}\left[\mathrm{By}\text{S.S.S.}\right]\\ \angle \mathrm{BDP}=\angle \mathrm{CDP}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \Rightarrow \mathrm{DP}\text{bisects\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BDC}\text{i.e.,}\angle \mathrm{D}.\\ \mathrm{Thus},\text{AP bisects}\angle \mathrm{A}\text{as well as}\angle \mathrm{D}.\\ \left(\mathrm{iv}\right)\angle \text{APB}=\angle \text{APC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{and}\angle \text{APB}+\angle \text{APC}=\text{180°}\\ \Rightarrow \angle \text{APB}+\angle \text{APB}=\text{180°}\\ \Rightarrow 2\angle \text{APB}=180\mathrm{°}\\ \Rightarrow \angle \text{APB}=\frac{180\mathrm{°}}{2}=90\mathrm{°}\\ \text{And BP}=\mathrm{C}\text{P}\\ \text{So},\text{ AP is perpendicular bisector of BC.}\end{array}$

Q.18 AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{AD}\text{is altitude in isosceles}\mathrm{\Delta }\text{ABC in which AB}=\text{AC.}\\ \mathrm{To}\text{\hspace{0.17em}prove}:\left(\text{i}\right)\text{AD bisects BC}\left(\text{ii}\right)\text{AD bisects\hspace{0.17em}\hspace{0.17em}}\angle \text{A}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\left(\mathrm{i}\right)\mathrm{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{ACD}\\ \text{AB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \angle \mathrm{ADB}=\angle \mathrm{ADC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Each 90°]}\\ \text{AD}=\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \mathrm{\Delta }\text{ABD}\cong \mathrm{\Delta }\text{ACD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \text{So},\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BD}=\text{CD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Therefore},\text{AD bisects BC.}\\ \left(\mathrm{ii}\right)\angle \mathrm{BAD}=\angle \mathrm{CAD}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Therefore},\text{AD bisects}\angle \mathrm{A}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.19

$\begin{array}{l}\mathrm{Two}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{BC}\mathrm{and}\mathrm{median}\mathrm{AM}\mathrm{of}\mathrm{one}\mathrm{triangle}\mathrm{ABC}\mathrm{are}\\ \mathrm{respectively}\mathrm{equal}\mathrm{to}\mathrm{sides}\mathrm{PQ}\mathrm{and}\mathrm{QR}\mathrm{and}\mathrm{median}\mathrm{PN}\mathrm{of}\\ \mathrm{\Delta PQR}(\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\mathrm{\Delta ABM}\cong \mathrm{\Delta PQN}\\ \left(\mathrm{ii}\right)\mathrm{\Delta ABC}\cong \mathrm{\Delta PQR}\end{array}$

Ans.

$\begin{array}{l}\text{Given:In}\Delta \text{ABC and}\Delta \text{PQR,}\text{AB}=\text{PQ,}\text{BC}=\text{QR and AM}=PN.\\ \mathrm{To}\text{prove:}\left(i\right)\Delta ABM\cong \Delta PQN\\ \left(ii\right)\Delta ABC\cong \Delta PQR\\ \text{Proof}:\text{(i)In}\mathrm{\Delta }\text{ABM and}\Delta \text{PQN}\\ \text{AB}=\text{PQ}\left[\mathrm{Given}\right]\\ \text{BM}=\text{QN}[\frac{1}{2}BC=\frac{1}{2}QR]\\ \text{AM}=PN\left[Given\right]\\ \therefore \Delta \text{ABC}\cong \Delta \text{PQR}\left[By\text{S}\text{.S}\text{.S}\text{.}\right]\\ \angle ABM=\angle PQN\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \Rightarrow \angle B=\angle Q\\ \left(ii\right)In\Delta \text{ABC and}\Delta \text{PQR}\\ \text{AB}=\text{PQ}\left[\text{Given}\right]\\ \angle B=\angle Q\left[\mathrm{Proved}\text{above}\right]\\ \text{BC}=\text{QR}\left[\text{Given}\right]\\ \therefore \Delta ABC\cong \Delta PQR\left[By\text{S}\text{.A}\text{.S}\text{.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}Hence Proved}\text{.}\end{array}$

Q.20 BE and CF are two equal altitudes of a triangle ABC.
Show that:
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e ABC is an isosceles triangle.

Ans.

$\begin{array}{l}\text{Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{ABC, BE}\perp \text{AC and CF}\perp \text{AB and BE = CF}\\ \text{To prove:}\mathrm{\Delta }\text{ABC is isosceles triangle.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:(i)\hspace{0.33em}}\mathrm{In}\text{}\mathrm{\Delta }\text{BCE and}\mathrm{\Delta }\text{CBF}\\ \angle \mathrm{BEC}=\angle \mathrm{CFB}\left[\mathrm{Each}\text{90°}\right]\\ \mathrm{BC}=\mathrm{BC}\left[\text{Common}\right]\\ \mathrm{BE}=\mathrm{CF}[\text{Given]}\\ \therefore \mathrm{\Delta }\text{BCE}\cong \mathrm{\Delta }\text{CBF\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \left(\mathrm{ii}\right)\mathrm{AB}=\mathrm{AC}[\mathrm{By}\mathrm{CPCT}]\\ \text{i.e ABC is an isosceles triangle}\end{array}$

Q.21 ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Ans.

$\begin{array}{l}\text{Given:In}\mathrm{\Delta }\text{ABC, AB}=\text{AC and AP}\perp \text{BC}\\ \text{To prove:}\mathrm{\Delta }\text{ABC is isosceles triangle.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:In}\mathrm{\Delta }\text{ABP and}\mathrm{\Delta A}\text{CP}\\ \angle \mathrm{APB}=\angle \mathrm{APC}\left[\mathrm{Each}\text{90°}\right]\\ \text{AB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \text{AP}=\text{AP\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \mathrm{\Delta }\text{ABP}\cong \mathrm{\Delta A}\text{CP\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \therefore \angle \mathrm{ABP}=\angle \mathrm{ACP}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \Rightarrow \angle \mathrm{B}=\angle \mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence Proved.}\end{array}$

Q.22 Show that in a right angled triangle, the hypotenuse is the longest side.

Ans.

$\begin{array}{l}\text{Given}:\mathrm{In}\mathrm{\Delta }\text{ABC,}\angle \text{B}=\text{90°}\\ \mathrm{To}\text{prove: AC is the longest side of}\mathrm{\Delta }\text{ABC}.\\ \text{Proof:In\hspace{0.17em}}\mathrm{\Delta }\text{ABC,}\\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ \angle \mathrm{A}+90\mathrm{°}+\angle \mathrm{C}=180\mathrm{°}\\ \angle \mathrm{A}+\angle \mathrm{C}=180\mathrm{°}-90\mathrm{°}\\ \angle \mathrm{A}+\angle \mathrm{C}=90\mathrm{°}\\ \text{Hence},\text{the other two angles have to be acute}(\text{i}.\text{e}.,\text{less than 9}0\mathrm{º}).\\ \therefore \angle \mathrm{B}\text{is the largest angle of the triangle.}\\ \angle \mathrm{B}>\angle \mathrm{A}\text{and}\angle \mathrm{B}>\angle \mathrm{C}\\ \therefore \mathrm{AC}>\mathrm{BC}\text{and AC}>\mathrm{A}\text{B}\\ \left[\begin{array}{l}\text{In any triangle},\\ \text{the side opposite to the larger}\left(\text{greater}\right)\text{angle is longer}.\end{array}\right]\\ \text{Therefore, AC is the largest side of the triangle.}\end{array}$

Q.23 In Fig. shown below, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Ans.

$\begin{array}{l}\text{Given: In}\mathrm{\Delta }\text{ABC, AB and AC are produced upto P and Q respectively.}\angle \text{PBC}<\angle \text{QCB.}\\ \text{To prove: AC}>\text{AB}\\ \text{Proof: Since,}\angle \text{PBC}+\angle \text{ABC}=\text{180°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{Pair of angles}\right]\\ \angle \text{PBC}=\text{180°}-\angle \text{ABC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{and}\angle \text{QCB}+\angle \text{ACB}=\text{180°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{Pair of angles}\right]\\ \angle \text{QCB}=\text{180°}-\angle \text{ACB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{It}\text{ is given that, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PBC}<\angle \text{QCB}\\ \text{180°}-\angle \text{ABC}<\text{180°}-\angle \text{ACB}\\ \Rightarrow -\angle \text{ABC}<-\angle \text{ACB}\\ \Rightarrow \angle \text{ACB}<\angle \text{ABC}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}<\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Side opposite to smaller}\\ \text{angle is smaller.}\end{array}\right]\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}>\text{AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.24

In Fig. shown, ∠B<∠A and ∠C<∠D. Show that AD
<bc< p=””></bc<>

Ans.

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{OAB,}\angle \text{B}<\angle \text{A and in}\mathrm{\Delta }\text{OCD,}\angle \text{C}<\angle \text{D}.\\ \mathrm{To}\text{prove:\hspace{0.17em}AD}<\text{BC}\\ \text{Proof}:\text{\hspace{0.17em}In}\mathrm{\Delta }\text{OAB,}\\ \angle \text{B}<\angle \text{A\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \Rightarrow \mathrm{OA}<\mathrm{OB}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Side}\text{opposite to smaller angle is}\\ \text{smaller.}\end{array}\right]\\ \mathrm{In}\text{}\mathrm{\Delta }\text{OCD,}\\ \angle \text{C}<\angle \text{D\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \Rightarrow \mathrm{OD}<\mathrm{OC}...\left(\mathrm{ii}\right)\left[\begin{array}{l}\text{Side opposite to smaller angle is}\\ \text{smaller.}\end{array}\right]\\ \text{Adding relation}\left(\mathrm{i}\right)\text{ and}\left(\mathrm{ii}\right)\text{, we get}\\ \mathrm{OA}+\mathrm{OD}<\mathrm{OB}+\mathrm{OC}\\ \Rightarrow \mathrm{AD}<\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence Proved.}\end{array}$

Q.25 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure below). Show that ∠A > ∠C and ∠B > ∠D.

Ans.

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{}\mathrm{quadrilateral}\text{ABCD,}\mathrm{A}\text{B is the smallest side and CD is the largest side.}\\ \mathrm{To}\text{prove:\hspace{0.17em}}\angle \text{A}>\angle \text{C and}\angle \mathrm{B}>\angle \mathrm{D}.\end{array}$

$\begin{array}{l}\text{Proof}:\text{\hspace{0.17em}In}\mathrm{\Delta }\text{ABC,}\\ \text{BC}>\text{AB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \angle \text{BAC}>\angle \text{ACB\hspace{0.17em}}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\end{array}$ $\begin{array}{l}\text{In}\mathrm{\Delta }\text{ACD,}\\ \text{CD}>\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \angle \text{CAD}>\angle \text{ACD\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \mathrm{Adding}\text{relation}\left(\mathrm{i}\right)\text{and relation}\left(\mathrm{ii}\right),\text{we get}\\ \angle \text{BAC}+\angle \text{CAD}>\angle \text{ACB}+\angle \text{ACD}\\ \angle \text{BAD}>\angle \text{BCD}\\ \Rightarrow \angle \text{A}>\angle \text{C\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{\hspace{0.17em}In}\mathrm{\Delta }\text{ABD,}\\ \text{AD}>\text{AB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \angle \text{ABD}>\angle \text{ADB\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iv}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \text{\hspace{0.17em}In}\mathrm{\Delta }\text{BCD,}\\ \text{CD}>\text{BC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \angle \text{CBD}>\angle \text{BDC\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{v}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \mathrm{Adding}\text{relation}\left(\mathrm{iv}\right)\text{and relation}\left(\mathrm{v}\right),\text{we get}\\ \angle \text{ABD}+\angle \text{CBD}>\angle \text{ADB}+\angle \text{BDC}\\ \angle \text{ABC}>\angle \text{ADC}\\ \Rightarrow \angle \text{B}>\angle \text{D\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{vi}\right)\\ \mathrm{From}\text{relation}\left(\mathrm{iii}\right)\text{ and relation}\left(\mathrm{vi}\right),\text{we have}\\ \angle \text{A}>\angle \text{C and}\angle \text{B}>\angle \text{D.\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.26 In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Ans.

$\begin{array}{l}\text{\hspace{0.17em}Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{PQR, PR > PQ and PS bisects}\angle \text{QPR.}\\ \text{To prove:\hspace{0.17em}}\angle \text{PSR}>\angle \text{PSQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof: In}\mathrm{\Delta }\text{PQR},\text{\hspace{0.17em}PS bisects}\angle \text{QPR}\\ \mathrm{i}.\mathrm{e}.,\angle \text{QPS}=\angle \text{RPS\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{In}\mathrm{\Delta }\text{PQS},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSR}=\angle \mathrm{PQS}+\angle \mathrm{QPS}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSR}=\angle \mathrm{PQR}+\angle \mathrm{QPS}\\ \Rightarrow \angle \mathrm{PQR}=\angle \mathrm{PSR}-\angle \mathrm{QPS}...\left(\mathrm{ii}\right)\\ \mathrm{In}\mathrm{\Delta }\text{PRS},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSQ}=\angle \mathrm{PRS}+\angle \mathrm{SPR}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSQ}=\angle \mathrm{PRQ}+\angle \mathrm{RPS}\\ \Rightarrow \angle \mathrm{PRQ}=\angle \mathrm{PSQ}-\angle \mathrm{RPS}\\ \Rightarrow \angle \mathrm{PRQ}=\angle \mathrm{PSQ}-\angle \mathrm{QPS}...\left(\mathrm{iii}\right)\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}PR}>\text{PQ}\\ \angle \mathrm{PQR}>\angle \mathrm{PRQ}\left[\begin{array}{l}\mathrm{Opposite}\text{angle of greater side}\\ \text{is greater.}\end{array}\right]\\ \angle \mathrm{PSR}-\angle \mathrm{QPS}>\angle \mathrm{PSQ}-\angle \mathrm{QPS}\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right)\right]\\ \angle \mathrm{PSR}>\angle \mathrm{PSQ}\mathrm{Hence}\text{proved.}\end{array}$

Q.27 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans.

$\begin{array}{l}\text{Given}:\text{A line l and a point A which is away from line l.}\\ \text{To prove: Perpendicular line segment from A to line l is the shortest.}\\ \text{Contruction: Draw AB}\perp \overleftrightarrow{\mathrm{l}},\text{AC and AD.}\\ \text{Proof: In}\mathrm{\Delta ABC},\angle \text{B=90°}\\ \mathrm{Then},\angle \text{A}+\angle \text{C}=\text{90°}\\ \Rightarrow \angle \mathrm{B}>\angle \mathrm{C}\\ \Rightarrow \mathrm{AC}>\mathrm{AB}\left[\mathrm{Side}\text{opposite to greater angle is greater.}\right]\\ \text{Similarly,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}>\angle \mathrm{D}\\ \Rightarrow \mathrm{AD}>\mathrm{AB}\left[\mathrm{Side}\text{opposite to greater angle is greater.}\right]\\ \text{Therefore},\text{it can be observed that of all line segments drawn}\\ \text{from a given point not on it},\text{the perpendicular line segment is}\\ \text{the shortest}.\mathrm{}\end{array}$

Q.28 ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of ΔABC.

Ans.

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.

In ∆ABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ∆ABC.

Q.29 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Ans.

A point in the interior of a triangle which is equidistant from all the sides is in centre. In centre is a point which is obtained by the intersection of angle bisectors.

Here, in ∆ABC, we can find the in centre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of ∆ABC.

Q.30 In a huge park, people are concentrated at three points (see figure below):

A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit. Where should an icecream parlour be set up so that maximum number of persons can approach it?

Ans.

Icecream parlour should be set up at equidistant point from three points A, B and C so that maximum number of people will approach there.

When we join three points A, B and C, we get a triangle. A point which is equidistant from three vertex is called Circumcentre.

In figure, Circumcentre O is obtained by intersection of perpendicular bisectors of sides of triangle ABC.

At this point of icecream parlour, maximum number of people will approach.

Q.31 Complete the hexagonal and star shaped Rangolies [see the figure below (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Ans.

\begin{array}{l}\left(i\right)Area\left(\Delta AOB\right)=\frac{\sqrt{3}}{4}{\left(side\right)}^2\\ =\frac{\sqrt{3}}{4}{\left(5\right)}^2\\ =\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\end{array}

\begin{array}{l}\text{Area of hexagonal – shaped}\\ \text{rangoli}=6\times Area\left(\Delta AOB\right)\\ =6\times \frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ =\frac{75\sqrt{3}}{2}{\text{cm}}^{\text{2}}\\ Areaof\text{equilateral triangle having its side as 1 cm}\\ =\frac{\sqrt{3}}{4}{\left(side\right)}^2\\ =\frac{\sqrt{3}}{4}{\left(1\right)}^2\\ =\frac{\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ Number\text{of equilateral triangles of 1 cm side that can be filled}\\ \text{in hexagonal-shaped rangoli}\\ =\frac{\left(\frac{75\sqrt{3}}{2}\right)}{\left(\frac{\sqrt{3}}{4}\right)}=150\end{array}

\begin{array}{l}\end{array} $\begin{array}{l}\text{Star}-\text{shaped rangoli has 12 equilateral triangles of side}\text{5 cm in it}.\\ \text{Area of equilateral triangle}\\ =\frac{\sqrt{3}}{4}{\left(side\right)}^2\\ =\frac{\sqrt{3}}{4}{\left(5\right)}^2\\ =\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{Area of star-shaped rangoli}\\ =12\times \frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ =75\sqrt{5}{\text{cm}}^{\text{2}}\\ \text{Area}of\text{equilateral triangle having its side as 1 cm}\\ =\frac{\sqrt{3}}{4}{\left(side\right)}^2\\ =\frac{\sqrt{3}}{4}{\left(1\right)}^2\\ =\frac{\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{Number of equilateral triangles of 1 cm side that can be filled}\\ \text{in hexagonal-shaped rangoli}\\ =\frac{\left(75\sqrt{3}{\text{cm}}^{\text{2}}\right)}{\left(\frac{\sqrt{3}}{4}\right)}=300\\ \text{Therefore},\text{star-shaped rangoli has more equilateral}\\ \text{triangles in it}.\end{array}$