Question

In a quadrilateral $ABCD,\angle B={90}^o$ and $\angle D={90}^o$, Prove that
$2{AC}^2-{AB}^2={BC}^2+{CD}^2+{DA}^2$

Answer 1

To prove:

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

Given:

A quadrilateral $ABCD$

$\angle B=\angle ABC={90}^{\circ }$

$\angle D=\angle CDA={90}^{\circ }$

Thus, side $AB||CD$ and $BC||AD$. This shows the given quadrilateral is a parallelogram.

Since, two opposite angles are right angles, the given quadrilateral is either a rectangle or a square.

Thus, the required quadrilateral is as shown in the figure below.

Now, apply the formula of pythagorus theorem.

In $△ABC$,

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

In $△CDA$,

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

Now, consider the LHS of the given expression.

$\Rightarrow 2A{C}^2-A{B}^2$

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

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

Now, consider the RHS of the given expression.

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

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

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

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

Thus,

$LHS=RHS$

Hence, the given expression is proved.