$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

(N/A) In $\triangle ADC$ and $\triangle BAC$:
$\angle ADC = \angle BAC$ (Given)
$\angle ACD = \angle BCA$ (Common angle)
Therefore,$\triangle ADC \sim \triangle BAC$ (By $AA$ similarity criterion).
We know that the corresponding sides of similar triangles are proportional.
Therefore,$\frac{CA}{CB} = \frac{CD}{CA}$.
Cross-multiplying gives $CA^2 = CB \cdot CD$.