If two medians in a triangle are equal in length,then the triangle is isosceles.
Let medians AM = BN in ∆ ABC. Extend each median to AP and BQ so that M and N are the midpoints of AP and BQ, respectively. Hence , AM = MP and BN = NQ. By the property of bisecting diagonals, ABPC and ABCQ are parallelograms. Hence CP and CQ are each parallel and equal to AB. We conclude that C lies on QP and C is the midpoint of QP.
Now , AM = BN , so 2AM = 2BN , hence AP = BQ. This shows that ABPQ is a trapezium with equal diagonals.
It is easy to see that such a trapezium is isosceles. One way to see this is to draw a line through A parallel to diagonal BQ, until it intersects line QP in point L. Thus, ABQL is a parallelogram, so ∠ALQ = ∠ABQ. On the other hand, ∆APL is isosceles since AL = BQ = AP; hence, ∠ ALQ = ∠ APQ.
Finally, AB || QP implies ∠ APQ = ∠ BAP. We conclude that ∠ BAP = ∠ ABQ, and ∆ ABQ and ∆ BAP are congruent by two equal sides and angles between these sides (SAS Criteria) . Therefore, BP = AQ and our trapezium is isosceles.
Hence ∠AQC=∠ BPC. Finally , ∆ACQ and ∆ BCP are congruent by AQ = BP, CQ = CP and ∠ AQC = ∠ BPC (SAS Criteria). We conclude that AC = BC and our original ABC is isosceles.