Point \( P \) on side \( BC \) of \( \triangle ABC \) is such that \( \angle APC = \angle BAC \). Demonstrate that \(AC^2 = BC \times PC\).
Proof:
Given a triangle \(ABC\) with a point \(P\) on \(BC\) such that \(\angle APC = \angle BAC\).
\(AC^2 = BC \cdot CP\)
Since \(\angle APC = \angle\)
By similarity criterion, \(\Delta APC \sim \Delta\) .
Since they have one common angle and \(\angle APC = \angle BAC\).
For similar triangles, the ratio of corresponding sides are equal.
\(\frac{AC}{BC} =\)
\(AC^2 = BC \cdot PC\)
Hence proved.