D is a point on the side BC of $\triangle ABC$ such that $\angle CAB = \angle CDA$. Show that $CA^2 = CB \times CD$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer
In △CAB and △CDA:
∠C = ∠C (common angle)
∠CAB = ∠CDA (given)
∴ △CAB ~ △CDA (AA similarity criterion)
Therefore, corresponding sides are proportional:
$$\frac{CA}{CD} = \frac{CB}{CA}$$
$$\Rightarrow CA^2 = CB \times CD \quad \textbf{(Proved)}$$
Source: Chapter 6 – Triangles, Section 6.4 Criteria for Similarity of Triangles
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Explanation
- The key step is identifying the correct pair of triangles: △CAB and △CDA (not △ABC and △ADC — note the vertex order matters for correspondence).
- ∠C is common to both triangles; the given condition provides the second equal angle → AA criterion applies.
- Once similarity is established, write the ratio with CA appearing in both numerator and denominator on opposite sides, then cross-multiply to get CA² = CB × CD.
- Examiners specifically check: (1) correct triangle correspondence in symbolic form, (2) reason for each equal angle, (3) correct ratio written before cross-multiplying.