So have a look at the figure below :
Let me list out the given information :
$1.$ $\angle ACB$ is a right angle
$2.$ $\Delta BCD$ is an equilateral triangle
To show that :
$\angle BAE = 2 \hspace{0.1cm} \angle EAC$
I have been struggling with this question that I came up with on my own while proving that we can indeed trisect any angle given to us.
I started by using the original method of angle chasing, but then I realized that we keep going round and round back to where we started.
Then I resorted to trigonometry and sine and cosine rule of triangles but in that, too, I kept returning back to the same point.
Any ideas on how I should prove the theorem ?