Question

$\angle BCD=\angle EDC$

Answer 1

Now in $\Delta ABC$ and $\Delta AED$

$AB=AE$ (given )

$\angle ABC=\angle AED$ (given )

And $BC=ED$ (given)

Hence $\Delta ABC\cong \Delta AED$ (By SAS rule)

So $AC=AD$ ......... $\left(1\right)$ (by CPCT)

And $\angle BCA=\angle EDA$ ......... $\left(2\right)$ (by CPCT)

Now in $\Delta CAD$, we know

$AC=AD$

So, from base angle theorem, we know

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

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

And $\angle EDC=\angle EDA+\angle ACD$

So the equation $\left(2\right)$ and $\left(3\right)$ we get

$\angle EDC=\angle BCA+\angle ACD$ ........... $\left(5\right)$

So from equation $\left(4\right)$ and $\left(5\right),$ we get

$\angle BCD=\angle EDC$