**Prove that the mid - point of the hypotenuse of a right triangle is equidistant from its vertices.**

let ABC be the right triangle right angled at b and AC is the hyp and D is the midpoint of hyp such that AD=CD

angle{ABC}=90degrees

Imagine triangle abc is inscribed in a cicle or semicircle such that AC is the diameter and so, angle{ABC}is angle in a semicircle which is always 90 degrees

Now,if AC=diametre, then AD=CD=radius

Also, BD=radius

**SO, AD=BD=CD WHICH WAS TO BE PROVED **

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