D is a point on side BC of a triangle ABC such that \(\angle ADC = \angle BAC\). Show that \(CA^2 = CB \cdot CD\).
Given: In \(\triangle ABC\), point D lies on BC such that \(\angle ADC = \angle BAC\).
To prove: \(CA^2 = CB \cdot CD\).
We know:
So we have two pairs of equal angles, hence
\[\triangle ADC \sim \triangle BAC \quad (\text{AA similarity}).\]
From the similarity, match the vertices:
\(D \leftrightarrow A,\ C \leftrightarrow C,\ A \leftrightarrow B.\)
Thus the corresponding sides are:
\[DC \leftrightarrow AC, \quad AC \leftrightarrow BC, \quad AD \leftrightarrow AB.\]
So we get the proportionality:
\[\frac{DC}{AC} = \frac{AC}{BC}.\]
Cross-multiplying,
\[DC \cdot BC = AC^2.\]
That is,
\[\boxed{CA^2 = CB \cdot CD}.\]
Hence proved.