Question

$AE$ is the bisector of $\angle CAD$. Also , $BA\parallel CE$ and $AB=AC$. Prove that $\angle EAC=\angle ACB$ and $ABCD$ is a parallelogram.

Answer 1

$ABCE$ is a parallelogram

$AB=AC\Rightarrow \angle ABC=\angle ACB$

Let \angle ABC = \angle ACB = x$

$\angle BAC=180°-2x$ [Angle sum property]

$AB\parallel CE\Rightarrow \angle BAC=\angle ACE$

$\angle BAC=\angle ACE=(180°-2x)$

$\angle BAC+\angle EAC+\angle EAD=180°$ [Linear Pair]

$(180°-2x)+2\angle EAC=180°$

$2\angle EAC=2x$

$\angle EAC=x$

Therefore, $\angle EAC=\angle ACB$

$\angle EAC=\angle ACB\Rightarrow AE\parallel BC$

Therefore, $ABCE$ is a parallelogram.