Question

In the given figure, ∠ACB = 90⁰ and CD ⊥ AB. Prove that

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

Answer 1

Given: ∠ACB = 90⁰ and CD ⊥ AB
To Prove: $\frac{{\mathrm{BC}}^2}{{\mathrm{AC}}^2}=\frac{\mathrm{BD}}{\mathrm{AD}}$
Proof:
In $△\mathrm{ACB}\mathrm{and}△\mathrm{CDB}$
$\angle \mathrm{ACB}=\angle \mathrm{CDB}={90}^{\circ }$ (Given)
$\angle \mathrm{ABC}=\angle \mathrm{CBD}$ (Common)
By AA similarity-criterion $△\mathrm{ACB}~△\mathrm{CDB}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$$\begin{gathered} \therefore \frac{\mathrm{BC}}{\mathrm{BD}}=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow {\mathrm{BC}}^2=\mathrm{BD}.\mathrm{AB}.....\left(1\right) \end{gathered}$$
In $△\mathrm{ACB}\mathrm{and}△\mathrm{ADC}$
$\angle \mathrm{ACB}=\angle \mathrm{ADC}={90}^{\circ }$ (Given)
$\angle \mathrm{CAB}=\angle \mathrm{DAC}$ (Common)
By AA similarity-criterion $△\mathrm{ACB}~△\mathrm{ADC}$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$$\begin{gathered} \therefore \frac{\mathrm{AC}}{\mathrm{AD}}=\frac{\mathrm{AB}}{\mathrm{AC}} \\ \Rightarrow {\mathrm{AC}}^2=\mathrm{AD}.\mathrm{AB}.....\left(2\right) \end{gathered}$$
Dividing (2) by (1), we get
$\frac{{\mathrm{BC}}^2}{{\mathrm{AC}}^2}=\frac{\mathrm{BD}}{\mathrm{AD}}$