D is a point on the side BC of $\displaystyle \triangle \mathrm{ABC}$ such that $\displaystyle \angle \mathrm{CAB}=\angle \mathrm{CDA}$. Show that $\displaystyle \mathrm{CA}^{2}=\mathrm{CB} \times \mathrm{CD}$.

\[\begin{aligned} & \triangle \mathrm{ADC} \sim \triangle \mathrm{BAC} \\ & \Rightarrow \frac{\mathrm{DC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{BC}} \\ & \Rightarrow \mathrm{AC}^{2}=\mathrm{DC} \times \mathrm{BC}^{\circ} \text { or } \mathrm{CA}^{2}=\mathrm{CB} \times \mathrm{CD} \end{aligned} \]