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

I a ABC, ...

Question

I a $△ A B C$ , $∠ A = 90^{o},$ CA=AB and D is a point on AB produced prove that
${D C}^{2} = {B D}^{2} = 2 A B \times A D$

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Solution

REF.Image.
In $△ A B C, ∠ A = 90$
$A C = A B$
D is an AB product
To prove : $D C^{2} - B D^{2} =$
$2 A B \times A D$
$\to$ Acc to Pythagoras theorem
In $△ A C D$
$C D^{2} = A C^{2} + A D^{.} . . (1)$
and In $△ A B C (∵ A B + B D = A D)$
$B C^{2} = A C^{2} + A B^{2} . . . (2)$
from (1):
$C D^{2} = A C^{2} + A D^{2}$
$C D^{2} - B D^{2} = A C^{2} + A D^{2} - B D^{2}$
$C D^{2} - B D^{2} = A C^{2} + A D^{2} - (A D - A B)^{2}$
$= A C^{2} + A D^{2} - A D^{2} - A B^{2} + 2 (A D / A B)$
$C D^{2} - B D^{2} = A C^{2} + (- A B^{2}) + 2 (A D) (A B)$
$= A B^{2} - A B^{2} + 2 (A B) (A D)$
or $(∵ A B = A C)$
$C D^{2} = B D^{2} = 2 A B \times A D$
Hence proved

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