$D$ and $E$ are points on the sides $C A$ and $C B$ repesctively of a $△ A B C$ , right angled at $C$ . Prove that $A E^{2} + B D^{2} = A B^{2} + D E^{2}$ .

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Solution

REF.Image
Construction: draw line ED.

In $Δ Δ A C B$ , applying Pythagoras theoram

$A B^{2} = A C^{2} + B C^{2}$ (1)

In $Δ A E C$ ,

$A E^{2} = A C^{2} + E C^{2}$ (2)

In $Δ B D C$

$B D^{2} = B C^{2} + C D^{2}$ (3)

In $Δ E C D$

$E D^{2} = C D^{2} + E C^{2}$ (4)

Adding (2) and (3)

$A E^{2} + B D^{2} = A C^{2} + E C^{2} + B C^{2} + C D^{2}$

$= A C^{2} + B C^{2} + E C^{2} + C D^{2}$

$A E^{2} + B D^{2} = A B^{2} + E D^{2}$

LHS=RHS

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

REF.ImageConstruction: draw line ED.In ΔΔACB, applying Pythagoras theoramAB2=AC2+BC2 ____________ (1)In ΔAEC,AE2=AC2+EC2 __________ (2)In ΔBDCBD2=BC2+CD2 ________ (3)In ΔECDED2=CD2+EC2 __________ (4)Adding (2) and (3)AE2+BD2=AC2+EC2+BC2+CD2=AC2+BC2+EC2+CD2AE2+BD2=AB2+ED2LHS=RHS

$D$ and $E$ are points on the sides $C A$ and $C B$ repesctively of a $△ A B C$ , right angled at $C$ . Prove that $A E^{2} + B D^{2} = A B^{2} + D E^{2}$ .