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 that $D$ is a point on side $BC$ of triangle $A BC$ such that $∠ A D C = ∠ B A C$ . We are to prove that $(C A)^{2} = CB \cdot C D$ . In triangles $A D C$ and $B A C$ , $∠ A D C = ∠ B A C$ (given) and $∠ C$ is common. By AA similarity, $△ A D C \sim △ B A C$ . By similarity, $\frac{C A}{CB} = \frac{C D}{C A}$ , which gives $(C A)^{2} = CB \cdot C D$ .

Step-by-step working

1
Draw the figure with triangle $A BC$ and point $D$ on $BC$ .
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2
In $△ A D C$ and $△ B A C$ , $∠ A D C = ∠ B A C$ (given) and $∠ C = ∠ C$ (common).
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3
By AA similarity criterion, $△ A D C \sim △ B A C$ .
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4
From similarity, $\frac{C A}{CB} = \frac{C D}{C A} \Rightarrow (C A)^{2} = CB \cdot C D$ .
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Concepts used

similar trianglesproveangle conditionproportionality

Approach

Step-by-step working

Concepts used

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