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

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Solution

Applying Pythagoras theorem in $Δ A C E$ , we obtain
$A C^{2} + C E^{2} = A E^{2} \dots \dots (1)$
Applying Pythagoras theorem in $Δ B C D$ , we obtain
$B C^{2} + C D^{2} = B D^{2} \dots \dots (2)$
Using equation (1) and equation (2), we obtain
$A C^{2} + C E^{2} + B C^{2} + C D^{2} = A E^{2} + B D^{2} \dots \dots (3)$
Applying Pythagoras theorem in $Δ C D E$ , we obtain
$D E^{2} = C D^{2} + C E^{2}$
Applying Pythagoras theorem in $Δ A B C$ , we obtain
$A B^{2} = A C^{2} + C B^{2}$
Putting the values in equation (3), we obtain
$D E^{2} + A B^{2} = A E^{2} + B D^{2}$

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D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2+BD2=AB2+DE2.

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