Experts please prove it !!!

Dear student,

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.

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