Question

BP = EP, where P is the point of intersection of BC and ED produced.

Answer 1

On $△ABC$ and $△AED,$

$AB=AE$

$\angle ABC=\angle AED$

and $BC=ED$

$\therefore △ABC\cong △AED$ (by SAS congruence rule.)

$AC=AD⟶\left(1\right)$ (By CPCT.)

and $\angle BCA=\angle EDA⟶\left(2\right)$ (By CPCT.)

In $△CAD,AC=AD⟶\left(from\right(1\left)\right)$

So, from base angle theorem,

$\angle ACD=\angle ADC⟶\left(3\right)$

$\angle BCD=\angle BCA+\angle ACD⟶\left(4\right)$

and $\angle EDC=angleEDA+\angle ADC\therefore \angle EDC=\angle BCA+\angle ACD⟶\left(5\right)⟹\angle BCD=\angle EDC$