The midpoint M of side AD of a convex quadrilateral ABCD is equidistant from all its vertices. Point K is a continuation of sides AB

The midpoint M of side AD of a convex quadrilateral ABCD is equidistant from all its vertices. Point K is a continuation of sides AB and CD. Prove that triangles KDA and KBC are similar.

Let us prove that ВK / AK = СK / DK
1) tr. ABM = tr. DСM (on both sides and the angle between them)
1.y. AMB = y.DСM (as vertical)
2.AM = MD by condition
3.ВM = CM by condition
2) tr. СMV-isosceles (MВ = MC by condition) hence y. MВС = yog. MСВ
3) angle СВK = 180 gr. – corner АВМ – corner МВС
y. ВСK = 180 gr. – y. MСD – ug. HСМ
hence ug.ВСK = ug.ВСK hence tr. ВСK-isosceles
4) ВK / AK = СK / DK hence tr. СВK is similar to tr. ADK