In the rhombus MPKH, the diagonals intersect at point O. Points A and B are taken on the sides MP and KH, respectively

| May 5, 2021 | education

In the rhombus MPKH, the diagonals intersect at point O. Points A and B are taken on the sides MP and KH, respectively, so that AM = KB. Find the sum of the angles MOA and BOH.

Let us prove that triangles MOA and BOK are equal.

By condition, AM = KB, OM = OK as halves of the diagonals of the rhombus, which are halved at point O.

Angle BKO = AMO as criss-crossing angles at the intersection of parallel lines AM and BK secant CM. Then the triangles MOA is equal to BOK on two sides and the angle between them.

Then the angle AOM = BOK, and since they are vertical angles, the OA segment is a continuation of the BO segment, and the BOA lie on one straight line and form a deployed angle of 180.

Angle AOH = 90, as formed at the intersection of the diagonals, then the angle MOA + BOH = 180 – AOH = 180 – 90 = 90.

Answer: The sum of the angles MOA and BOH is 90.



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