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

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Solution

Step 1: Form equations for $∆ A B E$ and $∆ D B C$ by Pythagoras theorem.
Given: $A B C$ is a right-angled triangle at $B$
In $∆ A B E$ use Pythagoras theorem
${(A E)}^{2} = {(A B)}^{2} + {(B E)}^{2} . . (i)$
In $∆ D B C$ use Pythagoras theorem
${(C D)}^{2} = {(B D)}^{2} + {(B C)}^{2} . . (i i)$
Add $(i)$ and $(i i)$
${(A E)}^{2} + {(C D)}^{2} = {(A B)}^{2} + {(B E)}^{2} + {(B D)}^{2} + {(B C)}^{2} {(A E)}^{2} + {(C D)}^{2} = {(A B)}^{2} + {(B C)}^{2} + {(B D)}^{2} + {(B E)}^{2} . . (i i i)$
Step 2: Prove the required condition
In $∆ A B C$ use Pythagoras theorem
${(A C)}^{2} = {(A B)}^{2} + {(B C)}^{2} . . . (i v)$
In $∆ D B E$ use Pythagoras theorem
${(D E)}^{2} = {(B E)}^{2} + {(B D)}^{2} . . . (v)$
Use $(i v), (v)$ in $(i i i)$
${(A E)}^{2} + {(C D)}^{2} = {(A C)}^{2} + {(D E)}^{2}$
Hence, the required condition ${(A E)}^{2} + {(C D)}^{2} = {(A C)}^{2} + {(D E)}^{2}$ is proved.

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

In the given figure $ △ABC$ is a right-angled triangle at $ B$.

Let $ D$ and $ E$ be any points on $ AB$ and $ BC$ respectively.

Prove that $ {\left(AE\right)}^{2} + {\left(CD\right)}^{2} = {\left(AC\right)}^{2} + {\left(DE\right)}^{2}$

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