P is a point on the side BC of $\triangle ABC$ such that $\angle APC = \angle BAC$. Prove that $AC^2 = BC \cdot CP$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
In △BAC and △APC:
∠BAC = ∠APC (Given)
∠ACB = ∠ACB (Common angle)
∴ △BAC ~ △APC (AA similarity criterion)
Therefore, their corresponding sides are proportional:
$$\frac{BC}{AC} = \frac{AC}{CP}$$
$$\Rightarrow AC^2 = BC \cdot CP \quad \textbf{(Proved)}$$
Source: Triangles, Section 6.4 (AA similarity criterion)
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Explanation
- The key is identifying the two triangles to compare: △BAC and △APC. Both share ∠C, and ∠BAC = ∠APC is given — that gives AA similarity.
- Once similarity is established, write the correct ratio of corresponding sides. Make sure the correspondence is right: B↔A, A↔A, C↔P (i.e., BC corresponds to AC, and AC corresponds to CP).
- The result $AC^2 = BC \cdot CP$ follows directly from the cross-multiplication of the proportion.
- This is essentially the same logic as Exercise 6.3, Q.13 ("D is a point on BC … show that CA² = CB·CD").