On the lateral sides of the isosceles triangle ABC with the base of the AC, equal segments AM and CN are laid.

On the lateral sides of the isosceles triangle ABC with the base of the AC, equal segments AM and CN are laid. ВD median of triangle ABC-intersects segment МN at point O. Prove that BO-median of triangle MBN.

Consider a triangle ABC. Since it is isosceles and the median BD is drawn to the AC, it means that BD is the bisector, median and height => angle ABD = angle CBD. Consider the triangle MBN. Since AM = CN and AB = BC, then MB = BN => triangle MBN is isosceles. BO is a bisector, since angle ABD = angle CBD (from what was proved above). Since BO is a bisector drawn to the base of an isosceles triangle, it is also the height and the median => BО – the median of the MBN triangle.