Question

In the adjacent figure, $AC=AE,AB=AD$ and $\angle BAD=\angle EAC$. Show that $BC=DE$.

Answer 1

It is given that $\angle BAD=\angle EAC$

Add $\angle DAC$ on both side we get

$\angle CAB=\angle DAE$

In $△BACand△EAD$

$AC=AE$ (Given)

$\angle CAB=\angle DAE$ (Given)

$AB=AD$ (Given)

$△BAC\cong △EAD$

$BC=DE$ (By corresponding portions of congruent triangles )