Question

In the given figure, AB = AC. Prove that DECB is an isosceles trapezium.

Answer 1

Here, AB = AC

`=>` ∠B = ∠C

∴ DECB is a cyclic quadrilateral

(In a triangle, angles opposite to equal sides are equal)

Also, ∠B + ∠DEC = 180°     ...(1)

(Pair of opposite angles in a cyclic quadrilateral are supplementary)

`=>` ∠C + ∠DEC = 180°   ...[From (1)]

But this is the sum of interior angles

On one side of a transversal.

∴ DE || BC

But ∠ADE = ∠AED = ∠C  ...[Corresponding angles]

Thus, ∠ADE = ∠AED

`=>` AD = AE

`=>` AB – AD = AC – AE  ...(∴ AB = AC)

`=>` BD = CE

Thus, we have, DE || BC and BD = CE

Hence, DECB is an isosceles trapezium