In the given figure, D is a point on the side BC of $\triangle ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CD \cdot CB$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer
In △CAD and △ABC:
∠ADC = ∠BAC (given)
∠ACD = ∠ACB (common angle)
∴ △CAD ~ △ABC (AA similarity criterion)
Therefore, corresponding sides are proportional:
$$\frac{CA}{CB} = \frac{CD}{CA}$$
$$\Rightarrow CA^2 = CD \cdot CB \quad \textbf{(Hence proved)}$$
Source: Triangles, Section 6.4 (Exercise 6.3, Q.13)
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Explanation
- The key step is identifying the correct pair of similar triangles: △CAD ~ △ABC (not △ADC ~ △BAC — vertex order matters for correct correspondence).
- Common angle at C links the two triangles; the given equal angles (∠ADC = ∠BAC) provide the second angle → AA criterion applies.
- Write the ratio with CA as the side common to both triangles on the same position in the correspondence, so cross-multiplying gives CA² directly.
- Examiners award 1 mark for proving similarity and 1 mark for the correct proportional sides leading to the result.