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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

In the given ...

Question

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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Solution

Step 1: Define the Pythagoras theorem based on given information
It is given that $△ A B C$ is a right-angled triangle at $B$ .
i.e., $∠ B = 90 °$
And, $D$ and $E$ be any points on $A B$ and $B C$ respectively.
Now, we know that, according to the Pythagoras theorem, in a right-angled triangle,
${(Hypotenuse)}^{2} = {(Perpendicular)}^{2} + {(Base)}^{2}$
Step 2: Apply the Pythagoras theorem in different triangles
By the Pythagoras theorem in $△ A B C$ ,
${(A C)}^{2} = {(A B)}^{2} + {(B C)}^{2}$ $. . . (i)$
By the Pythagoras theorem in $△ A B E$ ,
${(A E)}^{2} = {(A B)}^{2} + {(B E)}^{2}$ $. . . (ii)$
By the Pythagoras theorem in $△ C B D$ ,
${(C D)}^{2} = {(C B)}^{2} + {(B D)}^{2}$ $. . . (iii)$
By the Pythagoras theorem in $△ D B E$ ,
${(D E)}^{2} = {(D B)}^{2} + {(B E)}^{2}$ $. . . (iv)$
Step 3: Determine the required equation
On adding the equation $(ii)$ and equation $(iii)$ , we get,
${(A E)}^{2} + {(C D)}^{2} = {(A B)}^{2} + {(B E)}^{2} + {(C B)}^{2} + {(B D)}^{2}$
$\Rightarrow$ ${(A E)}^{2} + {(C D)}^{2} = {(A B)}^{2} + {(C B)}^{2} + {(B D)}^{2} + {(B E)}^{2}$
$\Rightarrow$ ${(A E)}^{2} + {(C D)}^{2} = {(A B)}^{2} + {(B C)}^{2} + {(D B)}^{2} + {(B E)}^{2}$ $[∵ C B = B C, B D = D B]$
$\Rightarrow$ ${(A E)}^{2} + {(C D)}^{2} = {(A C)}^{2} + {(D E)}^{2}$ [from the equations $(i)$ and $(iv)$ ]
Hence, it is proved that ${(A E)}^{2} + {(C D)}^{2} = {(A C)}^{2} + {(D E)}^{2}$ .

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Similar questions

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