Question

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

Answer 1

Given triangle ABC is a right angle triangle at B

Draw BD perpendicular to AC

In $\Delta ABC$ and $\Delta ADB$

$\angle ABC=\angle ADB$

$\angle A=\angle A$

$\therefore \Delta ABC\sim \Delta ADB$

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

$\Rightarrow A{B}^2=AC\times AD$.............................................(1)

In$\Delta ABC$ and $\Delta BDC$

$\angle ABC=\angle bDC$

$\angle C=\angle C$

$\therefore \Delta ABC\sim \Delta BDC$

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

$\Rightarrow B{C}^2=AC\times DC$

Add (1) and (2) we get

$A{B}^2+B{C}^2=AC\times AD+AC\times DC$

$=AC(AD+DC)=AC\times AC$

$=A{C}^2$

Hence proved.