Question

ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that

1. ∠A = ∠B
2. ∠C = ∠D
3. ΔABC ≅ ΔBAD
4. diagonal AC = diagonal BD

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

Answer 1

We have given a trapezium ABCD in which AB || CD and AD = BC.

(i) Produce AB to E and draw CE || AD       ...(1)

∵ AB || DC ⇒ AE || DC

Also, AD || CE                 ...[From (1)]

∴ AECD is a parallelogram.

⇒ AD = CE              ...(1)     ...[∵ Opposite sides of the parallelogram are equal]

But AD = BC         ...(2)   ...[Given]

By (1) and (2), BC = CE

Now, in ΔBCE, we have BC = CE

⇒ ∠CEB = ∠CBE        ...(3)      ...[∵ Angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180°       ...(4)   ...[Linear pair]

and ∠A + ∠CEB = 180°           ...(5)      ...[Co-interior angles of a parallelogram ADCE]

From (4) and (5), we get

∠ABC + ∠CBE = ∠A + ∠CEB

∠ABC = ∠A         ...[From (3)]

∠B = ∠A             ...(6)

(ii) AB || CD and AD is a transversal.

∴ ∠A + ∠D = 180°               ...(7)     ...[Co-interior angles]

Similarly, ∠B + ∠C = 180°           ...(8)

From (7) and (8), we get

∠A + ∠D = ∠B + ∠C

∠C = ∠D                          ...[From (6)]

(iii) In ΔABC and ΔBAD, we have

AB = BA             ...[Common]

BC = AD           ...[Given]

∠ABC = ∠BAD    ...[Proved]

∴ ΔABC ≅ ΔBAD      ...[By SAS congruency]

(iv) Since, ΔABC ≅ ΔBAD      ...[Proved]

AC = BD           ...[By C.P.C.T.]