Solve this

Solution:

Given: ∠ACB = 900 and CD ⊥ AB

To Prove: $\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}$

Proof:

In $\triangle \mathrm{ACB}$ and $\triangle \mathrm{CDB}$

$\angle \mathrm{ACB}=\angle \mathrm{CDB}=90^{\circ}$ (Given)

$\angle \mathrm{ABC}=\angle \mathrm{CBD}$ (Common)

By AA similarity-criterion $\triangle \mathrm{ACB} \sim \triangle \mathrm{CDB}$

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.

$\therefore \frac{\mathrm{BC}}{\mathrm{BD}}=\frac{\mathrm{AB}}{\mathrm{BC}}$

$\Rightarrow \mathrm{BC}^{2}=\mathrm{BD} \cdot \mathrm{AB} \quad \ldots .(1)$

In $\triangle \mathrm{ACB}$ and $\triangle \mathrm{ADC}$

$\angle \mathrm{ACB}=\angle \mathrm{ADC}=90^{\circ}$ (Given)

$\angle \mathrm{CAB}=\angle \mathrm{DAC}$ (Common)

By AA similarity-criterion $\triangle \mathrm{ACB} \sim \triangle \mathrm{ADC}$

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.

$\therefore \frac{\mathrm{AC}}{\mathrm{AD}}=\frac{\mathrm{AB}}{\mathrm{AC}}$

$\Rightarrow \mathrm{AC}^{2}=\mathrm{AD} \cdot \mathrm{AB} \quad \ldots \ldots(2)$

Dividing (2) by (1), we get

$\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}$