ABCD is a trapezium in which AB || CD and AD = BC. Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) Δ ABC ≅ Δ BAD

(iv) diagonal AC = diagonal BD

Steps of construction :

1. Extend AB.

2. Draw a line through C parallel to DA intersecting AB produced at E.

Given :

ABCD is a trapezium, AB || CD and AD = BC

By construction,

AD || CE

and

AE || DC.

In AECD, both pair of opposite sides are parallel, hence, AECD is a parallelogram

(i) From figure,

AD = CE (Opposite sides of parallelogram are equal)

As,

AD = BC

∴ BC = CE

We know that,

Angles opposite to equal sides in a triangle are also equal.

⇒ ∠CEB = ∠CBE ………(1)

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

We know that,

Sum of co-interior angles = 180°

⇒ ∠BAD + ∠CEB = 180°

⇒ ∠BAD + ∠CBE = 180° [From equation (1)] ….. (2)

From figure,

⇒ ∠ABC + ∠CBE = 180° (Linear pairs) ……. (3)

From equations (2) and (3), we get :

⇒ ∠BAD + ∠CBE = ∠ABC + ∠CBE

⇒ ∠BAD = ∠ABC.

∴ ∠A = ∠B

Hence, proved that ∠A = ∠B.

(ii) From figure,

AB || CD

We know that,

Sum of co-interior angles equal to 180°.

⇒ ∠A + ∠D = 180° ………(4)

⇒ ∠C + ∠B = 180° ………(5)

From equation (4) and (5), we get :

⇒ ∠A + ∠D = ∠C + ∠B

Since, ∠A = ∠B

We can say that,

⇒ ∠D = ∠C

Hence, proved that ∠C = ∠D.

(iii) Join AC and BD.

In ∆ ABC and ∆ BAD,

⇒ AB = BA (Common side)

⇒ BC = AD (Given)

⇒ ∠B = ∠A (above)

∴ ∆ ABC ≅ ∆ BAD (By S.A.S. congruence rule)

Hence, proved that Δ ABC ≅ Δ BAD.

(iv) Since ∆ ABC ≅ ∆ BAD,

We know that,

Corresponding parts of congruent triangles are equal.

⇒ AC = BD (By C.P.C.T.)

Hence, proved that diagonal AC = diagonal BD.