# Double Angle in Circumscribed Triangle

In triangle $ABC$, angle $ACB$ is $50$ degrees, and angle $CBA$ is $70$ degrees. Let $D$ be the foot of the perpendicular from $A$ to $BC$, $O$ the center of the circle circumscribed around triangle $ABC$, and $E$ the other end of the diameter which goes through $A$. Find the angle $DAE$, in degrees.

Hello,

I managed to solve the problem above using my own method, but I'm having trouble understanding the "official" solution.

This solution states that: Since triangle $ACD$ is right, $\angle CAD = 90 - \angle ACD = 90- 50 = 40$. Also, $\angle AOC = 2*\angle ABC = 2*70 = 140$. Since triangle $ACO$ is isosceles with $AO = CO$, $\angle CAO = (180 - \angle AOC)/2 = (180 - 140)/2 = 20$. Hence, $\angle DAE = \angle CAD - \angle CAO = 40 - 20 = 20$.

Most of this makes sense to me, but I don't understand why the solution states that $\angle AOC = 2*\angle ABC$.

I would really appreciate an explanation of how $\angle AOC$ is twice as large as $\angle ABC$.

Thanks ^_^