Question

$\text{In the given figure,}AB||EC,AB=ACandAEbisects\angle \text{DAC.Prove that:}$

$\left(i\right)\angle EAC=\angle ACB\left(ii\right)\text{ABCE is a parallelogram.}$

Answer 1

Since $ED$ bisects $\angle DAC$, therefore,

$\Rightarrow \angle DAE=\angle EAC$...(i)

Also, as $AB||EC$, therefore

$\Rightarrow \angle BAC=\angle ECA$ (Alternate interior angles) ...(ii)

Also, as $AB=AC$, therefore, $\angle ABC=\angle ACB$...(iii)

By Angle sum property, in $\Delta ABC$

$\Rightarrow \angle BAC={180}^{\circ }-2\angle ACB$ [From (iii)] ...(iv)

Also, $BD$ is a straight line, therefore,

$\Rightarrow \angle BAC={180}^{\circ }-2\angle EAC$ [From (i)] ...(v)

From (iv) and (v), we get,

$\Rightarrow \angle ACB=\angle EAC$ [Hence proved the first part]

But they also form a pair of alternate interior angles.

Hence, $AE||BC$

Therefore, $ABCE$ is a parallelogram as opposite pairs are parallel.