Question

ABCD is a trapezium in which ab parallel to dc and ad=bc. if ce is drawn parallel to ad, meeting ab at e. Prove that-(i) aecd is a parallelogram. (ii) ad=ec. (iii) triangle ceb is a isoceles triangle.​

Answer 1

Given:

Trapezium = ABCD

AB || DC and AD = BC.

To Find: -(i) aecd is a parallelogram. (ii) ad=ec. (iii) ΔCEB is an isoceles triangle.​

Solution:

Extending AB and drawing a line through C || AD, intersecting AE at point E, thus AECD is a parallelogram.

(i) AD = CE ( Parallelogram AECD opposite sides)

As, AD = BC (Given)

Therefore, BC = CE

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

Now,

∠A + ∠CEB = 180º ( Same side transversal angles )

∠A + ∠CBE = 180º ( As ∠CEB = ∠CBE) --- eq 1

As, ∠B + ∠CBE = 180º (Linear pair angles) --- eq 2

From equation 1 and 2

∠A = ∠B

(ii) AB || CD

∠A + ∠D = 180º ( Same side transversal angles )

∠C + ∠B = 180° ( Same side transversal angles )

Therefore,

∠A + ∠D = ∠C + ∠B

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

Therefore, ∠C = ∠D

(iii) In ΔABC and ΔBAD,

BC = AD (Given)

AB = BA (Common side)

∠B = ∠A (Proved before)

Therefore, ΔABC ≅ ΔBAD ( By SAS congruence rule)