In the middle of an isosceles triangle ABC (AB = AC), a point K is designated so that BK = CK.

In the middle of an isosceles triangle ABC (AB = AC), a point K is designated so that BK = CK. Prove that line AK is perpendicular to BC.

Since the triangle ABC is isosceles, the angle ABC = ACB.

Consider triangles ABK and ACK.

AB = AB, angle ABK = ACK since triangle ABC is isosceles, BK = CK by condition.

Then the triangles ABK and ACK are equal on two sides and the angle between them, which means the angle AKB = AKC. Angle AKB and angle KAC are adjacent angles, the sum of which is 180.

Then AKB + КАC = 180, 2 * AKB = 180.

The angle AKB = AKC = 180/2 = 90, which means that the AK is perpendicular to the BC, which was required to be proved.