Question

In the adjoining figure, AB || DC and ∠C = ∠D. Prove that:

1. AD = BC
2. AC = BD.

Answer 1

Given: AB || DC and ∠C = ∠D

To Prove:

1. AD = BC 
2. AC = BD

Proof [Step-wise]:

1. Because AB || DC, co‑interior (same‑side interior) angles along transversals give ∠A + ∠D = 180° and ∠B + ∠C = 180°.

Hence ∠A + ∠D = ∠B + ∠C.

2. Given ∠C = ∠D, subtracting from the equality in (1) yields ∠A = ∠B.

So the top base angles are equal.

3. Consider triangles ΔACD and ΔCBA.

∠ACD = ∠CAB   ...(Alternate interior angles, AC is a transversal of the parallel lines AB and DC)

∠ADC = ∠CBA   ...(Because ∠CBA = ∠BCD alternate interior and given ∠BCD = ∠ADC)

AC = CA   ...(Common side)

Therefore ΔACD ≅ ΔCBA by AAS two angles and the corresponding side. 

From the congruence, corresponding sides AD and BC are equal.

Thus, AD = BC. (Proved (i))

4. Now consider triangles ΔABD and ΔBAC.

AD = BC   ...(From step 3)

∠A = ∠B   ...(From step 2)

AB = BA   ...(Common side)

Therefore, ΔABD ≅ ΔBAC by SAS.

From the congruence, corresponding sides BD and AC are equal.

Thus, AC = BD. (Proved (ii))

(i) AD = BC and (ii) AC = BD, as required.