# 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 ^_^