The basic idea behind this proof is that in two right triangles if the hypotenuses are congruent and two legs are congruent then the triangles are congruent.
A different approach involves the fact that the centroid divides any median in the ratio 2:1. Suppose in triangle ABC, the medians AD and BE are equal and intersect at centroid G. Can you use the fact above to prove something about triangles AGE and BGD ?
posted by MATNTRNG
1 atriangleABC has a base BC and two rays AB andAC. midpoints of AB=M of AC =N
2 BN = CM given
3 MN is parallel to BC ( similar triangles )
4 arcs are drawn at B and C of equal lenghts ( median lenght)
5 a line parallel to BC is drawn above BC meeting arcs at M and N
6 MNCB is a trapezoid
7BN and CM are the trapezoid diagonals which are equal 4 above
8 MNCB is a regular trapezoid
9 a regular trapezoid is a truncated isosceles triangle
BM and CM extended meet at A the apex of isoscelesABC
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