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:
Given that \( \angle APC = \angle BAC \), we compare triangles \( \triangle APC \) and \( \triangle BAC \).

Using the Law of Sines in triangles \( \triangle APC \) and \( \triangle BAC \):
In \( \triangle APC \):
\[ \frac{AC}{\sin \angle APC} = \frac{CP}{\sin \angle CAP} \quad \text{(1)} \] In \( \triangle ABC \):
\[ \frac{AC}{\sin \angle BAC} = \frac{BC}{\sin \angle ABC} \quad \text{(2)} \]
Now, since \( \angle APC = \angle BAC \), the corresponding angles in the two triangles are equal.
Also, if we consider triangles \( \triangle APC \) and \( \triangle BAC \) and use a geometric similarity argument:

Let us draw altitude from \( A \) perpendicular to line \( BC \), and apply trigonometry or use triangle area approach.
But the more direct approach is to use the **Angle-Angle Similarity**:
From the given, \( \angle APC = \angle BAC \), and both triangles share angle \( \angle A \), we get:
\[ \triangle APC \sim \triangle CAB \quad \text{(by AA similarity)} \]
From similarity of triangles \( \triangle APC \sim \triangle CAB \):
\[ \frac{AC}{BC} = \frac{CP}{AC} \Rightarrow AC^2 = BC \cdot CP \]
Hence Proved:
\[ \boxed{AC^2 = BC \cdot CP} \]