Question

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
Prove that : $A{E}^2+B{D}^2=A{B}^2+D{E}^2$

Answer 1

Given:- $△ABC$, right angled at $C$.

Two points $D$ and $E$ are on the sides $AC$ and $AB$.

To prove:- ${AE}^2+{BD}^2={AB}^2+{DE}^2$

Construction:- Join the point $D$ with $B$ and $E$ and point $E$ with $A$.

Proof:-

Using pythagoras theorem,

${\text{Hypotenuse}}^2={\text{Perpendicular}}^2+{\text{Base}}^2$

In $△ACE$,

${AE}^2={AC}^2+{CE}^2.....\left(1\right)$

In $△BCD$,

${BD}^2={DC}^2+{BC}^2.....\left(2\right)$

In $△ABC$,

${AB}^2={AC}^2+{BC}^2.....\left(3\right)$

In $△DCE$,

${DE}^2={DC}^2+{CE}^2.....\left(4\right)$

Now,

${AE}^2+{BD}^2={AB}^2+{DE}^2$

${AC}^2+{CE}^2+{DC}^2+{BC}^2={AC}^2+{BC}^2+{DC}^2+{CE}^2$

L.H.S. $=$ R.H.S.

Hence proved.