Question

D is a point on side BC of $△ A BC$ such that $∠ A D C = ∠ B A C$ . Prove that $(C A)^{2} = CB \cdot C D$ .

Approach

Given $∠ A D C = ∠ B A C$ and $∠ C$ common, triangles $△ A D C$ and $△ B A C$ are similar by AA criterion. From similarity, $\frac{C A}{CB} = \frac{C D}{C A}$ , so $(C A)^{2} = CB \cdot C D$ .

Step-by-step working

1
In $△ A D C$ and $△ B A C$ , $∠ A D C = ∠ B A C$ (given), $∠ C = ∠ C$ (common).
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2
By AA similarity, $△ A D C \sim △ B A C$ .
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3
Corresponding sides are proportional: $\frac{C A}{CB} = \frac{C D}{C A}$ .
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4
Cross-multiply: $(C A)^{2} = CB \cdot C D$ .
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Concepts used

trianglessimilar trianglesproveproportionalityangle-angle similarity

Approach

Step-by-step working

Concepts used

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