In figure, if $\triangle \mathrm{ABE}$

Solution:

In figure,

$\Delta \mathrm{ABE} \cong \triangle \mathrm{ACD}$(Given)

$\Rightarrow \mathrm{AB}=\mathrm{AC}$ and $\mathrm{AE}=\mathrm{AD} \quad(\mathrm{CPCT})$

$\Rightarrow \frac{A B}{A C}=1$ and $\frac{A D}{A E}=1$

$\Rightarrow \frac{A B}{A C}=\frac{A D}{A E} \quad(E a c h=1)$

Now, in $\triangle \mathrm{ADE}$ and $\triangle \mathrm{ABC}$, we have

$\frac{A D}{A E}=\frac{A B}{A C}$ (proved)

i.e., $\frac{A D}{A B}=\frac{A E}{A C}$

and also $\angle \mathrm{DAE}=\angle \mathrm{BAC} \quad($ Each $=\angle \mathrm{A})$

$\Rightarrow \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$ (By SAS similarity criterion)