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

Applying Pythagoras theorem in $\Delta ACE$, we obtain
$A{C}^2+C{E}^2=A{E}^2\dots \dots \left(1\right)$
Applying Pythagoras theorem in $\Delta BCD$, we obtain
$B{C}^2+C{D}^2=B{D}^2\dots \dots \left(2\right)$
Using equation (1) and equation (2), we obtain
$A{C}^2+C{E}^2+B{C}^2+C{D}^2=A{E}^2+B{D}^2\dots \dots \left(3\right)$
Applying Pythagoras theorem in $\Delta CDE$, we obtain
$D{E}^2=C{D}^2+C{E}^2$
Applying Pythagoras theorem in $\Delta ABC$, we obtain
$A{B}^2=A{C}^2+C{B}^2$
Putting the values in equation (3), we obtain
$D{E}^2+A{B}^2=A{E}^2+B{D}^2$