prove that: (i)If all medians of a triangle are equal then triangle is equilateral.
                (ii)sum of any two medians of a triangle is greater than the third median.

Asked by araima2001 | 10th Sep, 2014, 08:36: PM

Expert Answer:

(i)
Let ABC be a triangle and our medians AD, BE and CF, where D is the mid-point of AB, E is the midpoint of AC and F is the midpoint of AB.
Let us say all the medians intersect at the point M, which would make M the centroid of our triangle ABC.
Now a centroid divides each median in the ratio 2:1. So we now have
AM:MD = BM:ME = CM:MF = 2:1
The question says that the medians are equal in length. So let us say AM = BM = CM = 2 units and MD = ME = MF = 1 unit.

Now consider the triangles, CDM and AFM. They can be proved to be congruent, because the CM = AM, FM = MD and the angle between AMF = angle between CMD. Now since these triangles are congruent, AF = CD.

But F is the mid-point of AB, so AF = FB. So we just proved AF = FB = BD = CD.
Hence, AF + FB = BD + CD. And hence, AB = BC.
You can prove similarly for the side AC and AB too and show that BC = AC, effectively proving that AB = BC = AC and hence its an equilateral triangle.

Answered by Prasenjit Paul | 11th Sep, 2014, 11:08: AM

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