TITLE: Prove that 3 points are in straight line.
QUESTION [8 upvotes]: I am having trouble finding a proper solution to this problem:
An equilateral triangle (ABC) is inscribed in a circle (o). Point D is in the shorter BC arc of circle o. Point E is symmetric to point B about line CD. Prove that points A, D, E are in the same straight line.
What I already tried:
In short - I noticed a rhombus and proved that ADE are in the same straight line, because the angle
$$
\measuredangle{ADE}=180^o
$$
My solution is wrong.
Regards,
Tom.

REPLY [2 votes]: Well, ∠CBA = ∠ACB = 60 deg. => arc AC = arc AB = 120, thus the big arc CB = 240, that means ∠CDB = 1/2 arc CB = 120, because it is an inscribed angle.