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using coordinate geometry prove mid point of hyptenuse is equidistant from the three vertices/
Asked by sanchita | 19 Mar, 2013, 09:03: PM

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/2b/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.

hence proved.

Answered by | 20 Mar, 2013, 05:04: AM

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