Question

In a quadrilateral $ABCD$, $\angle B=90°$, $AD²=AB²+BC²+CD²$. Prove that $\angle ACD=90°$.

Answer 1

Proving that $\angle ACD=90°$

Given: In quadrilateral $ABCD$

$\angle B=90°$

$AD²=AB²+BC²+CD²$

Construction: Join Diagonal $AC$

Consider $△ABC$

$△ABC$ is a right angled triangle at vertex $B$

By Pythagoras' theorem,

$A{B}^2+B{C}^2=A{C}^2$ $...\left(i\right)$

It is given that

$AD²=AB²+BC²+CD²$

From $\left(i\right)$

$\Rightarrow$$A{D}^2=A{C}^2+C{D}^2$$...\left(ii\right)$

By the converse of Pythagoras' theorem, if the sum of squares of two sides of triangle is equal to the square of the third side, then the triangle is a right angled triangle with the right angle at the common vertex to the two sides.

Consider $△ADC$

As, $A{D}^2=A{C}^2+C{D}^2$

$△ADC$ is a right angled triangle at vertex $C$

Hence, $\angle ACD=90°$.

Hence proved.