Question

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

Answer 1

Given:
$AC=AEAB=AD$

Also, $\angle BAD=\angle EAC$
On adding $\angle DAC$ on both the sides, we get
$\angle BAD+\angle DAC=\angle EAC+\angle DAC\Rightarrow \angle BAC=\angle DAE$

$\text{Consider}△ADE\text{and}△ABCAE=AC\text{(Given)}AD=AB\text{(Given)}\angle DAE=\angle BAC\text{(Proved above)}\text{By SAS congruence criterion,}△ADE\cong △ABC.$
$\text{By CPCT,}BC=DE.$