(A) D is a point on the side BC of a triangle ABC such that ∠ ADC =∠ BAC, prove that CA^2= CB. CD

In \(\displaystyle \triangle \mathrm{ABC}\), D is a point on side BC such that \(\displaystyle \angle \mathrm{ADC}=\angle \mathrm{BAC}\) In \(\displaystyle \Delta \mathrm{CBA}\) and \(\displaystyle \Delta \mathrm{CDA}\)

\[\begin{aligned} & \angle \mathrm{C}=\angle \mathrm{C} \text { (common) } \\ & \angle \mathrm{BAC}=\angle \mathrm{ADC} \text { (given) } \\ & \therefore \triangle \mathrm{CBA} \sim \triangle \mathrm{CAD} \text { (By AA similarity) } \\ & \therefore \text { their corresponding sides are proportional } \\ & =\frac{\mathrm{CA}}{\mathrm{CD}} \Rightarrow \mathrm{CA}^{2}=\mathrm{CB} \cdot \mathrm{CD} \end{aligned} \]

\(\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}\)

AD and AM are medians of \(\displaystyle \Delta \mathrm{ABC}\) and \(\displaystyle \Delta \mathrm{PQR}\) respectively.

\(\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}\)

\[\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}} \]

\(\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}\)

\(\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)

Also \(\displaystyle \angle \mathrm{B}=\angle \mathrm{Q} \quad(\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR})\)

\(\displaystyle \Rightarrow \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM}\) (SAS similarly)

\(\displaystyle \Rightarrow \frac{\boldsymbol{A} \boldsymbol{B}}{\boldsymbol{P} \boldsymbol{Q}}=\frac{\boldsymbol{A} \boldsymbol{D}}{\boldsymbol{P} \boldsymbol{M}}\)