D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$, prove that $CA^2 = CB \cdot CD$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:24 · grounding rag
Model Answer
To Prove: $CA^2 = CB \cdot CD$
Given: D is a point on BC of △ABC such that ∠ADC = ∠BAC.
Proof:
In △BAC and △ADC,
∠BAC = ∠ADC (Given)
∠BCA = ∠ACD (Common angle, i.e., ∠ACB = ∠DCA)
Therefore, △BAC ~ △ADC (AA similarity criterion)
Since the triangles are similar, their corresponding sides are in the same ratio:
$$\frac{CA}{CD} = \frac{CB}{CA}$$
Cross-multiplying:
$$CA \times CA = CB \times CD$$
$$\boxed{CA^2 = CB \cdot CD}$$
Hence proved.
Source: Chapter 6, Section 6.4 – Criteria for Similarity of Triangles
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Explanation
- Key step: Identify the two triangles to compare — △BAC and △ADC. They share ∠C and have ∠BAC = ∠ADC (given), so AA similarity applies immediately.
- Correspondence matters: Write the similarity as △BAC ~ △ADC carefully. A corresponds to A, B corresponds to D... no wait — match angles: ∠BAC ↔ ∠ADC and ∠BCA ↔ ∠DCA, so vertex C is common. The correct correspondence is △BAC ~ △ADC, giving $\frac{CA}{CD} = \frac{CB}{CA}$.
- Common error: Students often write the wrong correspondence and get an incorrect ratio. Always match vertices by equal angles.
- Final step: Cross-multiplication of the proportion directly gives $CA^2 = CB \cdot CD$.