Experts please prove it !!!
Let triangle ABC be a right triangle such that ∠BAC = 90°.... (1)
Let O be the mid-point of the hypotenuse BC.
Then, OB = OC.
Now, taking O as centre and OC as radius draw a circle passing through points B and C.
Let us suppose that this circle doesn't pass through point A.
Let it meet AB at A' or produced AB at A'.
∠BA'C = 90° [Angle in a semi-circle is 90°]
⇒ ∠BAC = ∠BA'C = 90° [using (1)] ... (2)
Equation (2), can't be possible because exterior angle of a triangle can never be equal to its corresponding interior angle.
So, ∠BAC = ∠BA'C = 90° is possible only when A and A' coincides with each other.
∴ the circle passing through B and C must pass through A.
⇒ OA = OB = OC [radius of the circle]
⇒ OA =
Hence, the line segment joining the mid point of the hypotenuse to the opposite vertex is half of the hypotenuse.