Question

In a quadrilateral $ABCD$,$\angle B=90°$, $A{D}^2=A{B}^2+B{C}^2+C{D}^2$, prove that $\angle ACD=90°$.

Answer 1

Prove of the given result $\angle ACD=90°$

$A{D}^2=A{B}^2+B{C}^2+C{D}^2.........\left(1\right)$

$\angle ABC=90°$

In $∆ABC$, using Pythagoras theorem, $\left[\because \angle \mathrm{ABC}=90°\right]$

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

Putting value in $\left(1\right)$ from $\left(2\right)$

$\Rightarrow A{D}^2=A{C}^2+C{D}^2$

$\therefore ∆ACD$follows Pythagoras theorem

So,$∆ACD$is a right angle triangle, right angled at $C$.

$\therefore \angle ACD=90°$

Hence,it is proved that $\angle ACD=90°$.