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

In $△ A BC$ and $△ D A C$ , $∠ B A C = ∠ A D C$ (given) and $∠ C$ is common. Therefore, $△ A BC \sim △ D A C$ by AA similarity. From similarity, $\frac{C A}{C D} = \frac{CB}{C A}$ , so $(C A)^{2} = CB \cdot C D$ .

Step-by-step working

1
In $△ A BC$ , $A B = A C \Rightarrow ∠ A BC = ∠ A CB$
1 mark
2
$∠ A D C = ∠ A D B = 9 0^{\circ}$ (given) $∴ △ A B D \sim △ ECF$ (AA)
0.5 marks
3
$\frac{A B}{EC} = \frac{B D}{CF} \Rightarrow \frac{A C}{EC} = \frac{B D}{CF}$
0.5 marks

Concepts used

similar trianglesangle-angle similarityproportionalityproof

Approach

Step-by-step working

Concepts used

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