> Hi, > > I've some troubles trying to prove this problem: > > Given a square (ABCD) , E is an internal point to the > square such that the angle ECD = angle EDC = 15°; > prove that ABE is an equilateral triangle. > > I've just managed to prove that ABE is isosceles, but > can't move forward. Hope someone can help me figure > out the right path! Hi GabMat, Bisect the square (side=a) with line FEG, F on CD, G on AB Let EF=h=(a/2)*tan15deg=(a/2)*(2-sqrt(3)) : EG=a-h=a-(a/2)*(2-sqrt(3))=(a/2)*sqrt(3) In triangleBEG BG=a/2, EG=(a/2)*sqrt(3) : BE=a, so triangleABE is equilateral. Regards, Peter Scales.