Question

ABC is a right angled triangle with $\angle C={90}^{\circ }$ and $CD\perp AB$ then prove that $\frac{{BC}^2}{{AC}^2}=\frac{BD}{AD}$.

Answer 1

Given triangle ABC angle C is right angle

Draw CD perpendicular on line AB

In $\Delta ACB$ and $\Delta BDC$

$\angle ACB=\angle BDC$

$\angle B=\angle B$

$\therefore \Delta ACB\sim \Delta BDC$

$\therefore \frac{BC}{BD}=\frac{AC}{CD}=\frac{AB}{BC}$

$\therefore B{C}^2=BD\times AB$ ........(1)

$\Delta ACB$ and $\Delta ADC$

$\angle ACB=\angle ADC$

$\angle A=\angle A$

$\therefore \Delta ACB\sim \Delta ADC$

$\therefore \frac{BC}{CD}=\frac{AC}{AD}=\frac{AB}{AC}$

$\therefore A{C}^2=AB\times AD$ ..........(2)

Divide (1) and (2) we get

$\frac{B{C}^2}{A{C}^2}=\frac{BD\times AB}{AB\times AD}=\frac{BD}{AD}$

Hence proved.