Question

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

Answer 1

$ABCD$ is a quadrilateral.

Produce $AB$ and $CD$ to meet at $E$. Also join $AC$ and $BD$.

In $△AED,$

$\angle A+\angle D={90}^o$ [ Given ]

$\Rightarrow$ $\angle E={180}^o-(\angle A+\angle D)$ [ Sum of angles of a triangle is ${180}^o.$ ]

$\Rightarrow$ $\angle E={180}^o-{90}^o$

$\therefore$ $\angle E={90}^o$

In right angled $△AED,$

$\Rightarrow$ $(AD{)}^2=(AE{)}^2+(DE{)}^2$ ---- ( 1 ) [ By Pythagoras theorem ]

In right angled $△BEC,$

$\Rightarrow$ $(BC{)}^2=(BE{)}^2+(EF{)}^2$ [ By Pythagoras theorem ]

On adding equation ( 1 ) and ( 2 ) we get,

$\Rightarrow$ $(AD{)}^2+(BC{)}^2=(AE{)}^2+(DE{)}^2+(BE{)}^2+(CE{)}^2$ ---- ( 3 )

In right angled $△AEC,$

$\Rightarrow$ $(AC{)}^2=(AE{)}^2+(CE{)}^2$ ---- ( 4 ) [ By Pythagoras theorem ]

In right angled $△BED,$

$\Rightarrow$ $(BD{)}^2=(BE{)}^2+(DE{)}^2$ ----( 5 )

Adding equation ( 4 ) and ( 5 ) we get,

$\Rightarrow$ $(AC{)}^2+(BD{)}^2=(AE{)}^2+(CE{)}^2+(BE{)}^2+(DE{)}^2$ ---- ( 6 )

From equation ( 3 ) and ( 6 ) we get,

$\Rightarrow$ $(AC{)}^2+(BD{)}^2=(AD{)}^2+(BC{)}^2$