Question

State true or false:

In a quadrilateral $ABCD,\angle B={90}^0$ and $\angle D={90}^0$, then
$2A{C}^2-A{B}^2=B{C}^2+C{D}^2+D{A}^2$

• A. True
• B. False

Answer 1

The correct option is A True

Diagonal $AC$ divides the quadrilateral in two equal half's.

From $△ABC;$ $A{C}^2=A{B}^2+B{C}^2⟶\left(1\right)$

From $△ADC;$ $A{C}^2=A{D}^2+D{C}^2⟶\left(2\right)$

Adding equating $\left(1\right)$ and $\left(2\right)$

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

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

Hence, the statement is true.