Question

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

Answer 1

consider triangles $ABC,ADE$

$AB=AD,AC=AE$

let angle $DAC=x$

angle$BAC=angleBAD+x$

$angleDAE=x+angleCAE=x+angleBAD=angleBAC$

so according to $SAS$ axiom triangles $ABC,ADE$ are congruent

so $BC=DE$