Question

P is a point on the side BC of $\triangle ABC$ such that $\angle APC = \angle BAC$. Prove that $AC^2 = BC \cdot CP$.

Hint:

Use angle-angle similarity to connect proportional sides and form product identities.

Answer 1

Given:
In triangle \( \triangle ABC \), point \( P \) lies on side \( BC \) such that:
\[\angle APC = \angle BAC\]
To Prove:
\[AC^2 = BC \cdot CP\]
Construction:
Join segments \( AP \) and \( AC \).

Proof:
Since \( \angle APC = \angle BAC \), we'll prove the statement by showing triangle similarity.

Using the Angle-Angle Similarity criterion:
From the given, \( \angle APC = \angle BAC \). Also, both triangles \( \triangle APC \) and \( \triangle BAC \) share angle \( \angle C \). Therefore:
\[\triangle APC \sim \triangle BAC \quad \text{(by AA similarity)}\]
From similarity of triangles \( \triangle APC \sim \triangle BAC \):
\[\frac{AC}{BC} = \frac{CP}{AC}\Rightarrow AC^2 = BC \cdot CP\]
Hence Proved:
\[\boxed{AC^2 = BC \cdot CP}\]