(a) In the given figure, Δ FEC ≅ GDB and ∠ 1=∠ 2. Prove that Δ ADE ∼ Δ ABC.

\(\displaystyle \Delta \mathrm{FEC} \cong \Delta \mathrm{GDB}\)

Therefore, \(\displaystyle \angle 3=\angle 4\)

In \(\displaystyle \Delta \mathrm{ABC}\),

\[\angle 3=\angle 4 \]

\[\therefore \mathrm{AB}=\mathrm{AC} \]

In \(\displaystyle \triangle \mathrm{ADE}, \quad \angle 1=\angle 2\)

\[A D=A E \]

Dividing (ii) by (i)

\[\begin{aligned} & \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}} \\ & \Rightarrow \mathrm{DE} \| \mathrm{BC} \\ & \angle 1=\angle 3 \text { and } \angle 2=\angle 4 \\ & \therefore \triangle \mathrm{ADE} \sim \triangle \mathrm{ABC} \end{aligned} \]

Produce AD to E such that \(\displaystyle \mathrm{AD}=\mathrm{DE}\) and join EC .

Produce PM to L such that PM = ML and join LR.

\[\begin{aligned} & \therefore \Delta \mathrm{ABD} \cong \Delta \mathrm{ECD} \\ & \therefore \mathrm{AB}=\mathrm{EC} \end{aligned} \]

Similarly, \(\displaystyle \mathrm{PQ}=\mathrm{LR}\)

\[\begin{aligned} & \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}} \\ & \frac{\mathrm{EC}}{\mathrm{LR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}=\frac{\mathrm{AE}}{\mathrm{PL}} \\ & \therefore \triangle \mathrm{AEC} \sim \triangle \mathrm{PLR} \end{aligned} \]

\[\Rightarrow \angle 2=\angle 4 \]

Similarly, \(\displaystyle \angle 1=\angle 3\)

Adding both, \(\displaystyle \angle \mathrm{BAC}=\angle \mathrm{QPR}\)

\[\therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \]