Question

Question 12
In a quadrilateral ABCD, $\angle A+\angle D={90}^{\circ }$. Prove that $A{C}^2+B{D}^2=A{D}^2+B{C}^2$.

Answer 1

Given, quadrilateral ABCD, in which
$\angle A+\angle D={90}^{\circ }$

To prove :
$A{C}^2+B{D}^2=A{D}^2+B{C}^2$

Construction : Produce AB and CD to meet at E.
Also, join AC and BD.

Proof :
In $\Delta AED$,
$\angle A+\angle D={90}^{\circ }\left[given\right]$
$\therefore \angle E={180}^{\circ }-(\angle A+\angle D)={90}^{\circ }$ $[becausesumofallanglesinatriangle={180}^{\circ }]$

Then, by using Pythagoras theorem,
$A{D}^2=A{E}^2+D{E}^2..\left(i\right)$

$In\Delta BEC$,
by using Pythagoras theorem,
$B{C}^2=B{E}^2+E{C}^2...\left(ii\right)$

On adding equations (i) and (ii), we get,
$A{D}^2+B{C}^2=A{E}^2+D{E}^2+B{E}^2+C{E}^2...\left(iii\right)$

$In\Delta AEC$,
by using Pythagoras theorem,
$A{C}^2=A{E}^2+C{E}^2...\left(iv\right)$

$\&inDeltaBED$,
by using Pythagoras theorem,
$B{D}^2=B{E}^2+D{E}^2...\left(v\right)$

On adding both equations, we get,
$A{C}^2+B{D}^2=A{E}^2+C{E}^2+B{E}^2+D{E}^2...\left(vi\right)$

From Eq. (iii) and (vi), we get,
$A{C}^2+B{D}^2=A{D}^2+B{C}^2$