#### Question

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

#### Solution

Here, AB = AC

⇒ ∠B =∠C

∴ DECB is a cyclic quadrilateral

(Ina 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

Is there an error in this question or solution?

Solution In the Following Figure Ab = Ac. Prove that Decb is an Isosceles Trapezium. Concept: Cyclic Properties.