$D$ 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

To prove:
$A E^{2} + B D^{2} = A B^{2} + D E^{2}$
Construction: Join $A E$ , $B D$ and $D E$
Proof:
In $△ A C E$ , by Pythagoras theorem
${(A E)}^{2} = {(A C)}^{2} + {(E C)}^{2} - - - (1)$
In $△ D C B$ , by Pythagoras theorem
${(B D)}^{2} = {(D C)}^{2} + {(B C)}^{2} . . . (2)$
In $△ A C B$ , by Pythagoras theorem
${(A B)}^{2} = {(A C)}^{2} + {(C B)}^{2} - - - (3)$
In $△ D C E$ , by Pythagoras theorem
${(E D)}^{2} = {(D C)}^{2} + {(C E)}^{2} - - - (4)$
Adding equation $(1)$ with equation $(2)$
${(A E)}^{2} + {(B D)}^{2} = {(A C)}^{2} + {(E C)}^{2} + {(D C)}^{2} + {(B C)}^{2} - - - - (5)$
From equation $(3)$ we have, ${(A B)}^{2} = {(A C)}^{2} + {(C B)}^{2}$
From equation $(4)$ we have, ${(E D)}^{2} = {(D C)}^{2} + {(C E)}^{2}$
Substitute equation $(3)$ and equation $(4)$ in equation $(5)$
${(A E)}^{2} + {(B D)}^{2} = {(A B)}^{2} + {(D E)}^{2}$
Hence, $A E^{2} + B D^{2} = A B^{2} + D E^{2}$

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Similar questions

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

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

D

E

C A

C B

△ A B C

C

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

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

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