In a regular triangle ABC on the side AB, we took point E and on the segment EC built a regular triangle

In a regular triangle ABC on the side AB, we took point E and on the segment EC built a regular triangle EKC towards point B. Prove that lines AC and BK are parallel.

A regular triangle is an equilateral triangle.
1) we will continue the segments ВK and AC to the left and to the right (so that ВK and AC become straight)
2) then EK is a secant (we will also continue it to the left and right so that EK becomes straight)
3) we prove that the angle BKE is equal to the angle EXC (where X is the point at the intersection of the straight line AC and EK)
4) Consider triangles AXE and BHK
1.Angle XEA = angle of the eyelid (as vertical)
2. Let us prove that the angle ХАЕ = angle ЕВК
1) angle ХАЕ = 180-angle ЕАС = 180-60 = 120 degrees
2) angle EBK = 60 + angle KVO (where point O is the point of intersection of straight lines EK and BC)
Find the KVO angle
To do this, we will prove that triangle AEC is equal to triangle СВK
consider the angles ACB and EСK, which are equal to 60 degrees (by the definition of an equilateral triangle)
we learn that angle ACE = angle BCK
(ACE = 60-ECB, BCK = 60-ECB)
AC side = BC side (tr. ABC equilateral by condition)
side of the EU = side of the COP (tr. EKС equilateral by condition)
hence the AEC triangle is equal to the SVK triangle on two sides and the angle between them)
This means that the angle EAC = angle KВO = 60 degrees
angle EBK = 60 + 60 = 120
hence the angle ХАЕ = ЕВК = 120 degrees
3) angle ВKE = 180-120-vertical angle ВEK
4) angle EXC = 180-120 – vertical angle XEA
hence the angle BKE is equal to the angle EXC
5) And it means that the angle BKE and the angle EXC are criss-crossing angles for two straight lines ВK and AC
6) hence the straight ВK and AC are parallel