Question

In given figure, $\angle ACB={90}^0$ and $CD\perp AB$. then prove that $\frac{{CB}^2}{{CA}^2}=\frac{BD}{AD}$

Answer 1

In triangles ACD and ABC, we have

$\angle ADC=\angle ACB$ [Each equal to ${90}^{\circ }$]

and, $\angle DAC=\angle BAC$ [Common]

So, by AA-criterion of similarity, we have

$△ACD\sim △ABC$

$\Rightarrow$ $\frac{AC}{AB}=\frac{AD}{AC}$

$\Rightarrow$ ${AC}^2=AB\times AD$......(i)

In ${△}^{\prime }sBCD$ and $BAC$, we have

$\angle BDC=\angle BCA$

$\angle DBC=\angle ABC$

So, by AA-criterion of similarity, we have

$BCD\sim BAC$

$\Rightarrow$ $\frac{BC}{BD}=\frac{BA}{BC}$

$\Rightarrow$ ${BC}^2=AB\times BD$ ......(ii)

Dividing (ii) by (i), we get

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

$\Rightarrow$ $\frac{{BC}^2}{{AC}^2}=\frac{BD}{AD}$ $\left[Henceproved\right]$