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.

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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 ^_^


Answer subtended by a chord to the centre is twice angle subtended from the chord to the circumference on the major arc. This is why angle AOC is twice as large as angle ABC because angle AOC is subtended by the chord AC to the centre and angle ABC is the angle subtended from the same chord onto the circumference and it falls on the major arc.


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