On the side AB of a regular triangle ABC, we took point M and on the segment MC

On the side AB of a regular triangle ABC, we took point M and on the segment MC on the same side as point B built a regular triangle MKC. Prove that AC and BK are parallel.

Consider triangles ABC and MKC – they are both regular, all angles of one and the second triangle are equal to 60 °.

This means that the MBC angle and the MKC angle are equal (60 ° each), they both rely on the MC segment, they can be inscribed in a circle (they will be inscribed angles based on the MC arc).

Points M, B, K and C lie on the same circle. This means that the CMC angle is equal to the CBC = 60 ° angle (as inscribed angles resting on the KC arc).

The angle MBK is equal to the sum of the angles MBK and CBK:

MBK = 60 ° + 60 ° = 120 °.

Angle A is equal to 60 ° (the angle of a regular triangle), in sum with the MBK angle it will give 180 °.

This means that angles A and MBK are internal one-sided with parallel AC and BK and secant AB.

Q.E.D.