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

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Solution

Given:
$△ A C B$ is a right angles triangle at C.
Construction:
Join $A E, B D$ and $E D$ .
To Prove:
$A E^{2} + B D^{2} = A B^{2} + D E^{2}$
Proof:
In $△ A C E$
By pythagoras theorem,
$A E^{2} = E C^{2} + A C^{2} . . . . . (1)$
In $△ B C D$
By pythagoras theorem,
$B D^{2} = B C^{2} + C D^{2} . . . . . (2)$
In $△ E C D$
By pythagoras theorem,
$E D^{2} = E C^{2} + D C^{2} . . . . . (3)$
In $△ A B C$
By pythagoras theorem,
$A B^{2} = B C^{2} + A C^{2} . . . . . (4)$
Adding 1) and 2) we get,
$A E^{2} + B D^{2} = E C^{2} + A C^{2} + B C^{2} + C D^{2} . . . . . . (5)$
From 3), 4) and 5) we get,
$A E^{2} + B D^{2} = A B^{2} + D E^{2}$ $[h e n c e p r o v e d]$

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