Question

In the given figure $AE=BC$ and $AE\left|\right|BC$ and the three sides $AB,CDandED$ are equal in length. If $\angle A=102°$, then find the measure of $\angle BCD$.

• A. $138°$
• B. $162°$
• C. $88°$
• D. $150°$

Answer 1

The correct option is B

$162°$

Finding $\mathbf{\angle }\mathit{B}\mathit{C}\mathit{D}$:

In the given figure

$AE=BC$ and $AE\parallel BC$

$\Rightarrow ABCE$ is a parallelogram. Therefore,

$AB=CE$ (opposite sides of a parallelogram) $...\left(1\right)$

$\angle BAE=\angle BCE=102°$ (opposite angles of a parallelogram) $...\left(2\right)$

$AB=CD=ED$ (given)

$\Rightarrow CE=CD=ED$ from $\left(1\right)$

As all the sides of the $△CDE$ are equal, so it is an equilateral triangle. Therefore,

$\angle ECD=\angle CDE=\angle DEC=60°$ (all the angles of an equilateral triangle are of $60°$.)

Now,

$$\begin{gathered} \angle BCD=\angle BCE+\angle ECD \\ \Rightarrow \angle BCD=102°+60° \\ \Rightarrow \angle BCD=162° \end{gathered}$$

Therefore the value of $\angle BCD={162}^{\circ }$.

Hence, the correct option is B.