Question

State and prove pythagorus theorum.

Answer 1

Pythagoras theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides

Given :

$△ABC$ right angles at $B$

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

Construction : Draw $DB\perp AC$

Proof: Since $BD\perp AC$

Therefore by AAS silimalrity criterion,

$△ADB\sim △ABC|△BDC\sim △ABC$

Since sides of similar triangles are in same ratio $|$ since similar triangle sides are in same ratio then

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

$AD.AC=A{B}^2------\left(1\right)$ $|$ $AC.CD=B{C}^2-------\left(2\right)$

Add $\left(1\right)$ and $\left(2\right)$ we get,

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

$AC(AD+DC)=A{B}^2+B{C}^2$

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

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

Hence proved