# ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ∆ABC ≅ ∆BAD

(iv) diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

**Solution:**

Let us join BD and AC in the figure tas shown below.. ADCE is a parallelogram.

**(i)** AD = CE (Opposite sides of parallelogram AECD are equal)

However, AD = BC (Given)

Therefore, BC = CE

∠CEB = ∠CBE (Angles opposite to equal sides in a triangle are also equal)

Consider, parallel lines AD and CE where AE is the transversal.

∠BAD + ∠CEB = 180° [Co-Interior angles]

∠BAD + ∠CBE = 180° ... (1) [Since, ∠CEB = ∠CBE]

However, ∠ABC + ∠CBE = 180° (Linear pair angles) ... (2)

From Equations (1) and (2), we see that

∠BAD = ∠ABC

Thus, ∠A = ∠B

**(ii)** AB || CD

∠A + ∠D = 180° (Angles on the same side of the transversal)

Also, ∠C + ∠B = 180° (Angles on the same side of the transversal)

∴ ∠A + ∠D = ∠C + ∠B

However, ∠A = ∠B [Using the result obtained in (i)]

∴ ∠C = ∠D

**(iii)** In ∆ABC and ∆BAD,

AB = BA (Common side)

BC = AD (Given)

∠B = ∠A (Proved before)

∴ ∆ABC ≅ ∆BAD (SAS congruence rule)

**(iv)** Since ∆ABC ≅ ∆BAD,

∴ AC = BD (By CPCT)

**Video Solution:**

## ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ∆ABC ≅ ∆BAD (iv) diagonal AC = diagonal BD [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

### NCERT Maths Solutions Class 9 - Chapter 8 Exercise 8.1 Question 12:

**Summary:**

If ABCD is a trapezium in which AB || CD and AD = BC, we have proved that ∠A = ∠B, ∠C = ∠D, △ABC ≅ △BAD by SAS congruence, and diagonal AC = diagonal BD.