Inside an equilateral triangle ABC, a point E is taken such that triangles ABE, BCE, ACE are equal.

Inside an equilateral triangle ABC, a point E is taken such that triangles ABE, BCE, ACE are equal. Find the angles for each of these triangles.

Consider triangles ∆ABE and ∆CBE: EB – common side, AB = CB – since ∆ABС is equilateral, therefore, the following angles are equal:

AEB = СEB, BAE = ВСЕ, ABE = CBE.

Now consider the triangle ∆AEC: side AE ​​is common with ∆ABE, side EC is common with ∆CBE, we get that the angles AEC = AEB = CEB are equal.

Because AEC + AEB + CEB = 360, then AEC = AEB = CEB = 360/3 = 120 °.

Because AE = EC, therefore ∆AEC is isosceles and the angles EAC and ECA are equal.

AEC + EAC + ECA = 180; EAC + ECA = 60; EAC = ECA = 30 °.

From the equality of the triangles ∆ABE, ∆ ВСЕ, ∆ ACE we have that:

EAC = ECA = EBA = BAE = EBC = ВСЕ = 30 °.

Answer: AEC = AEB = CEB = 120 °; EAC = ECA = EBA = BAE = EBC = ВСЕ = 30 °.