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

Point $D$ on side $BC$ of $△ A BC$ satisfies $∠ A D C = ∠ B A C$ . In $△ A D C$ and $△ B A C$ , $∠ A D C = ∠ B A C$ (given) and $∠ C$ is common, so $△ A D C \sim △ B A C$ by AA similarity. Hence $\frac{C A}{CB} = \frac{C D}{C A}$ , giving $(C A)^{2} = CB \cdot C D$ .

Step-by-step working

1
In $△ A D C$ and $△ B A C$ , $∠ A D C = ∠ B A C$ (given), $∠ C = ∠ C$ (common)
1 mark
2
$△ A D C \sim △ B A C$ by AA similarity criterion
0.5 marks
3
$\frac{C A}{CB} = \frac{C D}{C A} \Rightarrow (C A)^{2} = CB \cdot C D$
0.5 marks

Concepts used

trianglessimilar trianglesproveangle bisector

Approach

Step-by-step working

Concepts used

Similar question types