# NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1

Mathematics is an integral part of human thought and logic and is integral to students’ attempts to understand the world and themselves. As well as building mental discipline, it promotes logical reasoning and problem-solving abilities. Moreover, Mathematics is crucial to understanding other school subjects, such as Science, Social Science, and even Music and Art.

In order to determine quantitative solutions to experimental laws, scientists use the fundamentals of Mathematics. It is used in a variety of fields, including Finance, Medicine, Computer Science, Social Sciences, Natural Sciences, and Engineering. So, the concepts of Mathematics are incorporated into the curriculum of all subjects in higher education. Class 9 Mathematics introduces students to many essential concepts of the subject, which form the basis for higher education. To perform well in the examinations and be able to apply those concepts in their future studies, students in Class 9 should develop strong concepts in mathematics.

Chapter 7: Class 9 Mathematics is Triangles. In lower grades, students have seen a glimpse of this chapter. There is no substitute for diligent practice when it comes to Mathematics. NCERT structures the fundamentals of all the subjects. Therefore, Extramarks provides students with r so that they do not have to look elsewhere for properly detailed and easy solutions. Class 9 is an essential academic year in the academic career of students. Besides preparing students for their examinations, the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 provide them with the foundation for academic success. Students must develop comprehensive learning habits in Class 9, which helps them perform better in the Class 10 board examinations. In order to improve students’ academic performance and have a bright academic career, Extramarks provides them with Exercise 7.1 Class 9 Maths NCERT Solutions.

Multiple public and private schools in India are managed and controlled by the Central Board of Secondary Education. Most of the schools in the country follow this board of education, making it one of the most important boards of education in the country. By resolution of the Indian government, the CBSE Board was established in 1929. Schools affiliated with this educational board follow the NCERT curriculum. NCERT promotes and coordinates research related to school education and prepares and publishes model textbooks, supplementary materials, etc. Additionally, it serves as a clearinghouse for ideas and information related to school education, as well as a government agency dedicated to achieving universal elementary education goals. Founded in 1961 by the Government of India, the National Council of Educational Research and Training (NCERT) assists the Central and State Governments with educational policies and programmes.

## NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 Triangles

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### Access NCERT Answers for Class-9 Maths Chapter 7 – Triangles

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### NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.1) Exercise 7.1

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### Chapter wise NCERT Solutions for Class 9 Maths

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### Class 9 Maths Chapter 7 Includes:

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### NCERT Solutions for Class 9

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Q.1

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

Ans

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

Q.2

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

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:}\text{\hspace{0.17em}}\\ \left(i\right)\text{\hspace{0.17em}}\Delta \text{ABD}\cong \Delta B\text{AC}\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}BD=AC\\ \left(iii\right)\text{\hspace{0.17em}}\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}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle BAD=\angle ABC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Given}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AD=BC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Given}\right]\\ \therefore \Delta \text{ABD}\cong \Delta \text{BAC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{SAS}\right]\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}BD=AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{CPCT}\right]\\ \left(iii\right)\text{\hspace{0.17em}}\angle \text{ABD}=\angle \text{BAC}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\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

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

Q.4

$\begin{array}{l}\mathrm{l}\mathrm{and}\mathrm{m}\mathrm{are}\mathrm{two}\mathrm{parallell}\mathrm{ines}\mathrm{intersected}\mathrm{by}\mathrm{another}\mathrm{pair}\mathrm{of}\\ \mathrm{parallel}\mathrm{lines}\mathrm{p}\mathrm{and}\mathrm{q}\left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABC}\cong \mathrm{\Delta CDA}.\end{array}$

Ans

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

Q.5

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

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)\text{\hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}}\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 }\text{\hspace{0.17em}}\mathrm{ABQ}\mathrm{and}\text{\hspace{0.17em}}\mathrm{\Delta APB}\\ \angle \mathrm{BAQ}=\angle \mathrm{BAP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{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}}\left[\mathrm{Common}\right]\\ \angle \mathrm{AQB}=\angle \mathrm{APB}\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{each}\text{90°}\right]\\ \mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABQ}\cong \text{\hspace{0.17em}}\mathrm{\Delta APB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{​ AAS}\right]\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BQ}=\mathrm{BP}\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}}\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}}\mathrm{BP}=\mathrm{BQ}\end{array}$

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

Ans

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

Q.7

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

Ans

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

Q.8

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

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{In right}\mathrm{\Delta }\text{ABC,}\angle \text{C}=90\mathrm{°}\text{and M is mid-point on}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}hypotenuse AB,\hspace{0.17em}CM}=\text{DM.}\\ \text{To Prove:}\left(\mathrm{i}\right)\mathrm{\Delta AMC}\cong \mathrm{\Delta BMD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\angle \mathrm{DBC}\text{is a right angle.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{\Delta DBC}\cong \mathrm{\Delta ACB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iv}\right)\mathrm{CM}=\frac{1}{2}\mathrm{AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\left(\mathrm{i}\right)\text{In}\mathrm{\Delta AMC}\text{}\mathrm{and}\text{\hspace{0.17em}}\mathrm{\Delta BMD}\\ \mathrm{AM}=\mathrm{BM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AMC}=\angle \mathrm{BMD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Vertica}\mathrm{l}\text{opposite angles}\right]\\ \mathrm{CM}=\mathrm{DM}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Given]}\end{array}$

$\begin{array}{l}\therefore \mathrm{\Delta AMC}\cong \mathrm{\Delta BMD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{S}.\mathrm{A}.\mathrm{S}.\right]\\ \mathrm{AC}=\mathrm{DB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACM}=\angle \mathrm{BDM}\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}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACD}=\angle \mathrm{BDC}\\ ⇒\mathrm{AC}\parallel \mathrm{DB}\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}}\left[\mathrm{Alternate}\text{angles are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}+\angle \mathrm{DBC}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}}\left[\text{Co-interior angles}\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}}90\mathrm{°}+\angle \mathrm{DBC}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DBC}=180\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{°}\\ \mathrm{So},\text{\hspace{0.17em}}\angle \mathrm{DBC}\text{is a right angle.}\\ \left(\mathrm{iii}\right)\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta }\text{DCB and}\mathrm{\Delta ACB}\text{}\\ \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}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\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}}\left[\mathrm{Common}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCB}\cong \mathrm{\Delta ACB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{S}.\mathrm{A}.\mathrm{S}.\right]\\ \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]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{CM}=\mathrm{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}}\left[\mathrm{CM}=\mathrm{DM}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CM}=\frac{1}{2}\mathrm{AB}\end{array}$

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