Answer: \(\triangle BAC \sim \triangle ADC\) (AA test), so \(\frac{CA}{CD} = \frac{CB}{CA}\), i.e. \(CA^2 = CB \times CD\).
In \(\triangle BAC\) and \(\triangle ADC\):
- \(\angle BAC \cong \angle ADC\) (given)
- \(\angle BCA \cong \angle ACD\) (common angle C)
\(\therefore \triangle BAC \sim \triangle ADC\) (AA test), with B ↔ A, A ↔ D, C ↔ C.
Corresponding sides are in proportion:
\[\frac{BC}{AC} = \frac{AC}{DC}\] \[\therefore CA^2 = CB \times CD\]Hence proved.