(a) Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR. Show that Δ ABC ∼ Δ PQR.

In \(\displaystyle \Delta \mathrm{ABC}\) and \(\displaystyle \Delta \mathrm{PQR}\)

\[\begin{aligned} & \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M} \\ & \frac{A B}{P Q}=\frac{2 B D}{2 Q M}=\frac{A D}{P M} \end{aligned} \]

( ∵ D is midpoint of BC and M is midpoint of QR)

\[\begin{aligned} & \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \Rightarrow \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM} \\ & \Rightarrow \angle \mathrm{~B}=\angle \mathrm{Q}-\text { (i) } \end{aligned} \]

Now, In \(\displaystyle \triangle \mathrm{ABC}\) and \(\displaystyle \triangle \mathrm{PQR}\)

\[\begin{array}{ll} \frac{A B}{P Q}=\frac{B C}{Q R} & \text { (given) } \\ \angle \mathrm{B}=\angle \mathrm{Q} & \text { from (i) } \\ \therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} & \end{array} \]

In \(\displaystyle \Delta \mathrm{BMC}\) and \(\displaystyle \Delta \mathrm{EMD}\)

\[\begin{aligned} & \mathrm{MC}=\mathrm{MD} \\ & \angle \mathrm{CMB}=\angle \mathrm{EMD} \\ & \angle \mathrm{MBC}=\angle \mathrm{MED} \\ & \therefore \triangle \mathrm{BMC} \cong \triangle \mathrm{EMD} \\ & \Rightarrow \mathrm{BC}=\mathrm{DE} \end{aligned} \]

But AD = BC

\[\therefore \mathrm{AD}=\mathrm{DE} \]

\[\Rightarrow \mathrm{AE}=2 \mathrm{BC} \]

\(\displaystyle \Delta \mathrm{AEL} \sim \Delta \mathrm{CBL}\)

\[\therefore \frac{E L}{B L}=\frac{A E}{B C} \]

\[\Longrightarrow \frac{E L}{B L}=\frac{2 B C}{B C} \]

\[\Longrightarrow \frac{E L}{B L}=2 \]

\[\Rightarrow \mathrm{EL}=2 \mathrm{BL} \]