Question

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

Answer 1

Given:- A right angled triangle $ABC$, right angle at $B$

To prove:- $A{C}^2=A{B}^2+B{C}^2$

Proof:- Draw a perpendicular $BD$ from $B$ to $AC$

In $△ABC$ and $△ABD$

$\angle ADB=\angle ABC={90}^o$

$\angle DAB=\angle BAC$ ..... (Common angle)

$\therefore △ABC\sim △ABD$ ...... (Using AA similarity criteria)

Now, $\frac{AD}{AB}=\frac{AB}{AC}$ ..... (Corresponding sides are proportional)

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

Similarly, $△ABC\sim △BDC$

$\therefore B{C}^2=CD\times AC$ ...... $\left(ii\right)$

Adding equations $\left(i\right)$ and $\left(ii\right)$, we get

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

$\Rightarrow A{B}^2+B{C}^2=AC(AD+CD)$

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

$\Rightarrow A{B}^2+B{C}^2=A{C}^2$ $\left[henceproved\right]$