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Prove that the mid point of the hypotenu...

Prove that the mid point of the hypotenuse of a right triangle is equidistant from it vertices.

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Given : A `Delta` ABC in wchih `/_ BAC = 90^(@)` and `O` is the midpoint of `BC`
To Prove : `OA` `=` `OB` `=` `OC`
Construction : Draw a circle with `O` as centre and `OB` as radius.
Proof : Clearly, this circle with `O` as centre and `OB` as radius and this circle will pass through `B` as well as `C`.
If possible, suppose this circle does not through `A.` Let it meet `BA` or `BA` produced at A'.

The, `/_ BA'C = 90^(@)` [ angle in a semicircle ] .
But, `/_ BAC = 90^(@)` [given]
`:. /_ BA'C = /_ BAC`. .....(i )
But, this is wrong since an exterior angle of a triangle cannot be equal to its interior opposite angle.
This can be true only when `A'`coincides with A.
Thus, the above circle which passes through B and C must also pass through A.
`:. OA= OB = OC`.
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