In an isosceles triangle ABC, points K and M are the midpoints of the lateral sides AB and BC, respectively. BD is the median of the triangle. Prove that ΔВКD = ΔВМD.
Since the triangle is equilateral, the median BD is also the bisector of the angle ABC, and then the angle ABD = CBD.
In triangles BKD and BMD, side BD is common.
Points K and M are the midpoints of segments AB and BC, and since AB = BC, then BK = BM.
Then the triangles BKD and BMD are equal according to the first sign of equality of triangles – on two sides and the angle between them.
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