Question

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

Answer 1

Congruency in $\mathbf{△}\mathbf{BAC}\mathbf{,}\mathbf{△}\mathbf{DAE}$:

$\angle BAD=\angle EAC$ (given)

Add $\angle DAC$ to both sides

$$\begin{gathered} \angle BAD+\angle DAC=\angle EAC+\angle DAC \\ \Rightarrow \angle BAC=\angle DAE \end{gathered}$$

In $△BAC$ and $△DAE$

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

$AC=AE,AB=AD$ (given)

$\Rightarrow △BAC\cong △DAE$ (SAS criteria)

$BC=DE$ (by CPCT)

Hence $\mathbf{BC}\mathbf{=}\mathbf{DE}$ proved.