Question

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

Answer 1

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

$\angle BAD+\angle DAC=\angle EAC+\angle DAC$ [add $\angle DAC$ on both sides]

$\therefore \angle BAC=\angle DAE$

In $△BAC$ and $△DAE$

$AB=AD$ (Given)

$\angle BAC=\angle DAE$ (Proved above)

$AC=AE$ (Given)

$\therefore △BAC\cong △DAE$ (By SAS congruence rule)

$\therefore BC=DE$ (By CPCT)