Question

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

Answer 1

We are given that $\angle BAD=\angle EAC$
Adding $\angle DAC$ to both sides, we get
$\angle BAD+\angle DAC=\angle EAC+\angle DAC$
$\angle BAC=\angle DAE$ …(i)
Now in $\Delta BAC$ and $\Delta EAD$, we have
$\angle BAC=\angle DAE$ [using (i)]
$AB=AD$ (given)
$AC=AE$ (given)
$\Delta BAC\cong \Delta EAD$ (by SAS congruence rule)
$BC=DE$ (by c.p.c.t)
Hence proved.