Question

State true or false:

In a pentagon $ABCDE$, $AB=AE,BC=ED$ and $\angle ABC=\angle AED$, then

$BP=EP$, where $P$ is the point of intersection of $BC$ and $ED$ produced.

• A. True
• B. False

Answer 1

The correct option is A True

Given a pentagon $ABCDE$ such that

$AB=AE,$ $BC=ED$ $\xi$ $\angle ABC=\angle AED,$

Then $BD=ED,$ where D is the point of intersection of BC and ED

In $△ABC$ and $△AED$

$AB=AE$

$\angle ABC=\angle AED$

$BC=ED$

$\Rightarrow △ABC\cong △AED$ ..... Using SAS criterion

$\therefore AC=AD$ and $\angle BCA=\angle EDA.$

In $△CAD,$ $AC=AD$

From base angle theorem, $\angle ACD=\angle ADC$

$\angle BCD=\angle EDA+\angle ADC$

$\Rightarrow \angle BCD=\angle EDC$

$\therefore$ The statement is true.

Hence, the answer is true.