Question

In the following figure, AE and BC are equal and parallel and the three sides AB, CD, and DE are equal to one another. If angle A is 102^o. Find angles AEC and BCD.

Answer 1

In the given figure

Given that AE = BC

We have to find ∠AEC and ∠BCD

In the quadrilateral AECB

(AE = BC and AE || BC)

So quadrilateral is a parallelogram

⇒  ∠BAE + ∠AEC = 180°

⇒  102° + ∠AEC = 180°

⇒  ∠AEC = 78°

Also, ∠BAE = ∠BCE = 102°

AB = EC (because AECB is a parallelogram )

Now consider ΔECD

EC = ED = CD [Since AB = EC ]

Therefore ΔEDC is an equilateral triangle.

⇒ ∠ECD = 60°

∠BCD = ∠BCE + ∠ECD

⇒ ∠BCD = 102° + 60°

⇒ ∠BCD = 162°

Therefore ∠AEC = 78° and ∠BCD = 162°