using coordinate geometry prove mid point of hyptenuse is equidistant from the three vertices/
Let ABC be the triangle such that B lies at origin, and BC = a and BA = b.
thus, coordinates of B= (0,0), A = (0,b) and C = (a,0)
D is the midpoint of AC thus coordinates of D = (a/2, b/2)
by distance formula, AD = sqrt [a2/4 + (b/2 - b)2] = sqrt [a2/4 + b2/4]
CD = sqrt [b2/4 + (a/2 - a)2] = sqrt [a2/4 + b2/4]
Now, BD = sqrt [a2/4 + b2/4]
thus, AD = CD = BD.
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