In the ABCD square, point K is the middle of the BC side, point M is the seridine of the AB side.

In the ABCD square, point K is the middle of the BC side, point M is the seridine of the AB side. Prove that lines AK and MD are perpendicular, and triangles AEM (E is the intersection point of lines AK and MD) and ABK are similar.

Since ABCD is a square, then AB = BC = CD = AD.

According to the condition, the point K and M is the middle of the sides BC and AB, then ВK = AM.

Triangles AВK and ADM are rectangular in which ВK = AM, AB = AD, then the triangles are similar in two legs.

Then the angle ВAK = ADM.

In triangle AM, the sum of the angles АМD + АDM = 90, then in the triangle AME the sum of the angles MAE + AME = 90, which means the angle AEM = 180 – 90 = 90.

Then the segments DM and AK intersect at right angles, as required.

Triangles ABK and AEM are rectangular, for which the angle A is common, then the triangles are similar in acute angle, which was required to prove.