NCERT Solutions for Class 9 Maths Exercise 8.1 Chapter 8

NCERT Solutions for Class 9 Maths Exercise 8.1 Chapter 8 deal with the properties of a parallelogram and its diagonals. NCERT stands for National Council of Educational Research and Training. The chapter teaches what a quadrilateral is and how special quadrilaterals like the rectangle, rhombus and square are built from a parallelogram. This exercise uses triangle congruence to prove those properties step by step.

These questions ask you to prove three things: that a parallelogram with equal diagonals is a rectangle, that the diagonals of a square are equal and cut each other at right angles, and that a parallelogram whose diagonal bisects one angle is a rhombus. Each proof is written line by line in easy English, so students can revise before an exam, parents can check the steps at home, and teachers can use them in class. A free printable PDF of this exercise is also available for offline practice.

NCERT Solutions for Class 9 Maths Exercise 8.1 Chapter 8 – Questions and Answers

Q1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:
Given: ABCD is a parallelogram in which AC = BD.
To prove: ABCD is a rectangle.

In triangle ADC and triangle BCD,
AD = BC (opposite sides of a parallelogram are equal)
DC = CD (common side)
AC = BD (given)
So, triangle ADC is congruent to triangle BCD (by S.S.S., that is Side-Side-Side).
So, ∠ADC = ∠BCD (by C.P.C.T., that is Corresponding Parts of Congruent Triangles).

Now AD is parallel to BC and DC is the transversal.
So, ∠ADC + ∠BCD = 180° (co-interior angles).
∠ADC + ∠ADC = 180°
2∠ADC = 180°
∠ADC = 90°

A parallelogram with one angle of 90° is a rectangle.

Final answer: ABCD is a rectangle. Hence proved.

Q2. Show that the diagonals of a square are equal and bisect each other at right angles.

Answer:
Given: ABCD is a square. Its diagonals AC and BD meet at O.

(i) The diagonals are equal
In triangle ABC and triangle BAD,
AB = BA (common side)
∠ABC = ∠BAD = 90° (all angles of a square are right angles)
BC = AD (all sides of a square are equal)
So, triangle ABC is congruent to triangle BAD (by S.A.S., that is Side-Angle-Side).
So, AC = BD (by C.P.C.T.).

(ii) The diagonals bisect each other
In triangle AOB and triangle COD,
∠AOB = ∠COD (vertically opposite angles)
∠OAB = ∠OCD (alternate interior angles, as AB is parallel to DC)
AB = CD (all sides of a square are equal)
So, triangle AOB is congruent to triangle COD (by A.A.S., that is Angle-Angle-Side).
So, OA = OC and OB = OD (by C.P.C.T.).

(iii) The diagonals meet at right angles
In triangle AOB and triangle COB,
OA = OC (proved above)
AB = CB (all sides of a square are equal)
OB = OB (common side)
So, triangle AOB is congruent to triangle COB (by S.S.S.).
So, ∠AOB = ∠COB (by C.P.C.T.).
But ∠AOB + ∠COB = 180° (linear pair on line AC).
So, 2∠AOB = 180°, which gives ∠AOB = 90°.

Final answer: The diagonals of a square are equal, they bisect each other, and they meet at right angles. Hence proved.

Q3. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that (i) it bisects ∠C and (ii) ABCD is a rhombus.

Answer:
Given: ABCD is a parallelogram and AC bisects ∠A, so ∠BAC = ∠DAC.

(i) AC bisects ∠C
AB is parallel to DC and AC is the transversal.
So, ∠BAC = ∠DCA (alternate interior angles) … (1)
AD is parallel to BC and AC is the transversal.
So, ∠DAC = ∠BCA (alternate interior angles) … (2)
Given, ∠BAC = ∠DAC … (3)
From (1), (2) and (3), ∠DCA = ∠BCA.
So, AC divides ∠C into two equal parts.

(ii) ABCD is a rhombus
From (2) and (3), ∠BAC = ∠BCA.
In triangle ABC, sides opposite to equal angles are equal.
So, BC = AB.
In a parallelogram, opposite sides are equal, so AB = DC and BC = AD.
Therefore, AB = BC = CD = DA.

Final answer: AC bisects ∠C, and ABCD is a rhombus. Hence proved.

Q4. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer:
Given: ABCD is a quadrilateral. AC = BD, AC is perpendicular to BD, OA = OC and OB = OD.

In triangle AOB and triangle COB,
AO = OC (given)
∠AOB = ∠COB (each 90°)
OB = OB (common side)
So, triangle AOB is congruent to triangle COB (by S.A.S.).
So, AB = BC (by C.P.C.T.) … (i)

In triangle BOC and triangle DOC,
BO = OD (given)
∠BOC = ∠DOC (each 90°)
OC = OC (common side)
So, triangle BOC is congruent to triangle DOC (by S.A.S.).
So, BC = CD (by C.P.C.T.) … (ii)

In triangle COD and triangle AOD,
CO = OA (given)
∠COD = ∠AOD (each 90°)
OD = OD (common side)
So, triangle COD is congruent to triangle AOD (by S.A.S.).
So, CD = DA (by C.P.C.T.) … (iii)

From (i), (ii) and (iii), AB = BC = CD = DA.
All four sides are equal, and the diagonals bisect each other, so ABCD is a rhombus.

Now show one angle is 90°.
In triangle ABD and triangle BAC,
AD = BC (proved above)
AB = BA (common side)
BD = AC (given)
So, triangle ABD is congruent to triangle BAC (by S.S.S.).
So, ∠DAB = ∠CBA (by C.P.C.T.).
Since AD is parallel to BC, ∠DAB + ∠CBA = 180°.
So, 2∠DAB = 180°, which gives ∠DAB = 90°.

Final answer: All four sides are equal and one angle is 90°, so ABCD is a square. Hence proved.

Q5. ABCD is a trapezium in which AB ∥ CD and AD = BC. Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) triangle ABC ≅ triangle BAD (iv) diagonal AC = diagonal BD.

Answer:
Given: ABCD is a trapezium, AB is parallel to CD, and AD = BC.
Construction: Draw CE parallel to DA, meeting AB produced at E. Join AC.

AB is parallel to CD, so AE is parallel to DC.
Also CE is parallel to DA (by construction).
So AECD is a parallelogram, which gives AD = CE.
But AD = BC (given), so CE = BC.

(i) ∠A = ∠B
In triangle BCE, CE = BC, so it is an isosceles triangle.
So, ∠CBE = ∠CEB (angles opposite to equal sides are equal).
On the straight line AE, ∠ABC + ∠CBE = 180° (linear pair) … (1)
AD is parallel to CE and AE is the transversal, so ∠DAB + ∠CEB = 180° (co-interior angles) … (2)
Since ∠CBE = ∠CEB, from (1) and (2) we get ∠DAB = ∠ABC, that is ∠A = ∠B.

(ii) ∠C = ∠D
AB is parallel to DC, so ∠A + ∠D = 180° and ∠B + ∠C = 180°.
Since ∠A = ∠B, we get ∠C = ∠D.

(iii) Triangle ABC ≅ triangle BAD
In triangle ABC and triangle BAD,
AB = BA (common side)
∠ABC = ∠BAD (proved in part i)
BC = AD (given)
So, the triangles are congruent (by S.A.S.).

(iv) AC = BD
By C.P.C.T., diagonal AC = diagonal BD.

Final answer: ∠A = ∠B, ∠C = ∠D, triangle ABC is congruent to triangle BAD, and AC = BD. Hence proved.

Q6. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Answer:
Given: ABCD is a rhombus, so AB = BC = CD = DA.

Diagonal AC
In triangle ABC, AB = BC (sides of a rhombus are equal).
So, ∠BAC = ∠BCA (angles opposite to equal sides are equal) … (1)
AB is parallel to DC and AC is the transversal.
So, ∠BAC = ∠DCA (alternate interior angles) … (2)
From (1) and (2), ∠BCA = ∠DCA, so AC bisects ∠C.
AD is parallel to BC and AC is the transversal.
So, ∠DAC = ∠BCA (alternate interior angles) … (3)
From (1) and (3), ∠BAC = ∠DAC, so AC bisects ∠A.

Diagonal BD
In triangle ABD, AB = AD (sides of a rhombus are equal).
So, ∠ABD = ∠ADB … (4)
AB is parallel to DC and BD is the transversal.
So, ∠ABD = ∠BDC (alternate interior angles) … (5)
From (4) and (5), ∠ADB = ∠BDC, so BD bisects ∠D.
AD is parallel to BC and BD is the transversal.
So, ∠ADB = ∠DBC (alternate interior angles) … (6)
From (4) and (6), ∠ABD = ∠DBC, so BD bisects ∠B.

Final answer: AC bisects ∠A and ∠C, and BD bisects ∠B and ∠D. Hence proved.

Q7. In triangle ABC and triangle DEF, AB = DE, AB ∥ DE, BC = EF and BC ∥ EF. Vertices A, B and C are joined to vertices D, E and F. Show that (i) ABED is a parallelogram (ii) BEFC is a parallelogram (iii) AD ∥ CF and AD = CF (iv) ACFD is a parallelogram (v) AC = DF (vi) triangle ABC ≅ triangle DEF.

Answer:

(i) In quadrilateral ABED, AB = DE and AB is parallel to DE (given).
If one pair of opposite sides is equal and parallel, the quadrilateral is a parallelogram.
So, ABED is a parallelogram.

(ii) In quadrilateral BEFC, BC = EF and BC is parallel to EF (given).
By the same rule, BEFC is a parallelogram.

(iii) ABED is a parallelogram, so AD = BE and AD is parallel to BE … (1)
BEFC is a parallelogram, so BE = CF and BE is parallel to CF … (2)
From (1) and (2), AD = CF and AD is parallel to CF.

(iv) In quadrilateral ACFD, AD = CF and AD is parallel to CF (proved in iii).
So, ACFD is a parallelogram.

(v) In a parallelogram, opposite sides are equal.
So, AC = DF.

(vi) In triangle ABC and triangle DEF,
AB = DE (given)
BC = EF (given)
AC = DF (proved in v)
So, triangle ABC is congruent to triangle DEF (by S.S.S.).

Final answer: All six parts are proved.

Q8. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB.

Answer:
Given: In triangle ABC, ∠C = 90°, M is the mid-point of AB, and MD is parallel to BC.

(i) In triangle ABC, M is the mid-point of AB and MD is parallel to BC.
By the converse of the mid-point theorem, a line through the mid-point of one side and parallel to another side cuts the third side into two equal parts.
So, D is the mid-point of AC, that is AD = DC.

(ii) MD is parallel to BC and AC is the transversal.
So, ∠MDA = ∠BCA (corresponding angles).
But ∠BCA = 90°.
So, ∠MDA = 90°, which means MD is perpendicular to AC.

(iii) In triangle MDC and triangle MDA,
DC = DA (proved in part i)
∠MDC = ∠MDA (each 90°)
MD = MD (common side)
So, the triangles are congruent (by S.A.S.).
So, CM = MA (by C.P.C.T.).
M is the mid-point of AB, so MA = ½ AB.
Therefore, CM = MA = ½ AB.

Final answer: D is the mid-point of AC, MD is perpendicular to AC, and CM = MA = ½ AB. Hence proved.

Q9. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:
Given: In quadrilateral ABCD, the diagonals meet at O with AO = OC, BO = OD, and all four angles at O are 90°.

In triangle AOB and triangle COB,
AO = OC (given)
∠AOB = ∠COB (each 90°)
OB = OB (common side)
So, the triangles are congruent (by S.A.S.), which gives AB = BC … (i)

In triangle BOC and triangle DOC,
BO = OD (given)
∠BOC = ∠DOC (each 90°)
OC = OC (common side)
So, the triangles are congruent (by S.A.S.), which gives BC = DC … (ii)

In triangle COD and triangle AOD,
CO = OA (given)
∠COD = ∠AOD (each 90°)
OD = OD (common side)
So, the triangles are congruent (by S.A.S.), which gives DC = DA … (iii)

From (i), (ii) and (iii), AB = BC = CD = DA.
The diagonals also bisect each other, so ABCD is a parallelogram with all four sides equal.

Final answer: ABCD is a rhombus. Hence proved.

Q10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that (i) triangle APB ≅ triangle CQD (ii) AP = CQ.

Answer:
Given: ABCD is a parallelogram, AP is perpendicular to BD and CQ is perpendicular to BD.

(i) In triangle APB and triangle CQD,
∠APB = ∠CQD (each 90°)
∠ABP = ∠CDQ (alternate interior angles, as AB is parallel to DC and BD is the transversal)
AB = CD (opposite sides of a parallelogram)
So, triangle APB is congruent to triangle CQD (by A.A.S.).

(ii) By C.P.C.T., AP = CQ.

Final answer: Triangle APB is congruent to triangle CQD, and AP = CQ. Hence proved.

Q11. ABCD is a trapezium in which AB ∥ DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that F is the mid-point of BC.

Answer:
Given: AB is parallel to DC, E is the mid-point of AD, and EF is parallel to AB.
Let EF cut the diagonal BD at O.

In triangle ABD, E is the mid-point of AD and EO is parallel to AB.
By the converse of the mid-point theorem, O is the mid-point of BD, so DO = OB.

AB is parallel to DC and EF is parallel to AB, so EF is parallel to DC.
Therefore, OF is parallel to DC.

In triangle BCD, O is the mid-point of BD and OF is parallel to DC.
By the converse of the mid-point theorem, F is the mid-point of BC, so BF = CF.

Final answer: F is the mid-point of BC. Hence proved.

Q12. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Answer:
Let the four angles be 3x, 5x, 9x and 13x.
The sum of the angles of a quadrilateral is 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = 360° ÷ 30
x = 12°

First angle = 3 × 12° = 36°
Second angle = 5 × 12° = 60°
Third angle = 9 × 12° = 108°
Fourth angle = 13 × 12° = 156°

Check: 36° + 60° + 108° + 156° = 360°. Correct.

Final answer: 36°, 60°, 108° and 156°