# 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.