If two medians of a triangle are equal, prove that the triangle formed by a segment
of each median and the third side is an isosceles triangle�
Let ABC be a triangle with medians AD and BE such that AD = BE.
Let F be the midpoint of AB.
Draw FC and GC, where C is the midpoint of AC and G is the midpoint of BC.
Since F is the midpoint of AB, we have AF = FB.
Since D is the midpoint of BC, we have GD = GC.
Now consider triangle AFG and triangle BCG:
Both have angle G in common.
AF = FB and GD = GC.
By the midpoint theorem, FC = CG.
Therefore, by the side-angle-side (SAS) congruence, triangle AFG is congruent to triangle BCG.
In particular, this implies that AG = BG.
Now consider triangle ACG and triangle ABF:
Both have angle A in common.
AC = CB (since C is the midpoint of AC and D is the midpoint of BC).
AF = FB (since F is the midpoint of AB).
Therefore, by the side-angle-side (SAS) congruence, triangle ACG is congruent to triangle ABF.
In particular, this implies that CG = BF.
Finally, consider triangle BGF:
GB = AB/2 (since G is the midpoint of BC and F is the midpoint of AB).
AF = FB (since F is the midpoint of AB).
AG = BG (as proved above).
Therefore, by the side-side-side (SSS) congruence, triangle BGF is congruent to triangle AGF.
In particular, this implies that BF = CG and BG = AG.
Therefore, the triangle formed by a segment of each median and the third side is an isosceles triangle.