In given figure ABC is a right-angled triangle at B. Let D and E be any points on AB and BC respectively. Prove that ${A E}^{2} + {C D}^{2} = {A C}^{2} + {D E}^{2}$

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Solution

As $Δ A B E$ is right angled triangle, then,

$\Rightarrow$ ${(A E)}^{2} = {(A B)}^{2} + {(B E)}^{2}$ ............(1)

Similarly, $Δ D B C$ is right angled triangle, so,

$\Rightarrow$ ${(C D)}^{2} = {(B D)}^{2} + {(B C)}^{2}$ ............(2)

Add equation (1) and (2),

$\Rightarrow$ ${(A E)}^{2} + {(C D)}^{2} = (A B^{2} + B E^{2}) + (B D^{2} + B C^{2})$

$\Rightarrow$ ${(A E)}^{2} + {(C D)}^{2} = (A B^{2} + B C^{2}) + (B D^{2} + B E^{2})$ ............(3)

In $Δ A B C$ ,

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

In $Δ D B E$ ,

$\Rightarrow$ $D E^{2} = B E^{2} + B D^{2}$

So, equation (3) becomes,

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

$\Rightarrow$ ${(A E)}^{2} + {(C D)}^{2} = {(A C)}^{2} + {(D E)}^{2}$ . $[H e n c e p r o v e d]$

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