Question

State and prove Pythagoras theorem.

Answer 1

Pythagoras' theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

According to Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

Given that a right-angled triangle $ABC$ is right-angled at $B$

We need to prove that $A{C}^2=A{B}^2+B{C}^2$

Construction: Draw a perpendicular $BD$ meeting $AC$ at $D$.

We know that $∆ADB~∆ABC$

Therefore, $\raisebox{1ex}{$AD$}\!\left/ \!\raisebox{-1ex}{$AB$}\right.=\raisebox{1ex}{$AB$}\!\left/ \!\raisebox{-1ex}{$AC$}\right.$ (from corresponding sides of similar triangles)

Or,$A{B}^2=AD\times AC.....eq1$

Also,$∆BDC~∆ABC$

Therefore, $\raisebox{1ex}{$CD$}\!\left/ \!\raisebox{-1ex}{$BC$}\right.=\raisebox{1ex}{$BC$}\!\left/ \!\raisebox{-1ex}{$AC$}\right.$ (corresponding sides of similar triangles)

Or $B{C}^2=CD\times AC.....eq2$

Adding eq.$1$ and eq.$2$,

$$\begin{gathered} A{B}^2+B{C}^2=AD\times AC+CD\times AC \\ A{B}^2+B{C}^2=AC(AD+CD) \end{gathered}$$

Since,$AD+CD=AC$

Therefore,

$A{C}^2=A{B}^2+B{C}^2$

Hence, Pythagoras' theorem is proved.