Given ABCD parallelogram AD = 2AB AM-bisector of angle BAD. Prove AN = MN.
| December 7, 2020 | education
Presumably point N is the intersection point of the sectrices of angles A and B. The bisector in the parallelogram “cuts off” the isosceles triangle, that is, AB = BM, and point M divides the BС side in half. Similarly, with the bisector from the angle B, which will be the median in the isosceles triangle ABM, hence AH = NM.
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