In an isosceles triangle ABC with base AC, point K is marked on the median BD, and points M and N

In an isosceles triangle ABC with base AC, point K is marked on the median BD, and points M and N are marked on sides AB and BC, respectively. It is known that the angle BKM = angle BKN, angle BMK = 110 degrees. a) Find the angle BNK. b) Prove that lines MN and BK are mutually perpendicular.

Since the triangle ABC is isosceles, then its median BD is also the height and the bisector, then the angle ABD = CBD.

In the triangle BKM and BKН, the side BK is common, the angle BKM = ВKН by condition, the angle MBK = НBK by construction, then the triangles BKM and BKН are equal to the AO side and two adjacent angles, which means BM = BH, and the angle BKM = BMK = 110.

Since AB = BC, and BM = BH, then the triangles ABC and BHM are similar in two proportional sides and the angle between them, then the angle BAC = BMH, and so these are the corresponding angles, then MH is parallel to AC, and then BD is perpendicular to AC, which and it was required to prove.