Suppose $C$ is any point on a circle, above a diameter $AB$. $P$ and $Q$ are points on the minor arcs $\widehat{AC}$ and $\widehat{BC}$. Prove that $$\angle APC + \angle CQB = \frac32\pi$$ Currently I drew the shape $APCQB$ to be a pentagon inside the circle and let $\angle APC =\alpha, \angle CQB=\beta$. Tried using the diameter to create right angles to generate an equation using angle sum = $3\pi$, tried introducing cyclic quads too, the diagram became too full and I was not getting anywhere. Any help please?

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    $\begingroup$ If $C$ is on the diameter, what is meant by "arcs $\widehat{AC}$ and $\widehat{BC}$"? $\endgroup$ – Nick Matteo May 19 '14 at 13:39
  • $\begingroup$ AB is the diameter, I drew C on top of the diameter, P to the left of C and Q to the right of C $\endgroup$ – user145591 May 19 '14 at 21:42

Suppose $O$ is the center of the circle.

An inscribed angle on a circle, from some point on the major arc between two points, is half the size of the central angle to the same two points (see, e.g. Wikipedia.) On the other hand, an inscribed angle from a point in the minor arc between two points is the supplement of this, i.e. $\pi$ minus half the central angle. (see e.g. here.) For instance, $\angle APC = \pi - \frac 1 2 \angle AOC$. Also, $\angle CQB = \pi - \frac 1 2 \angle COB$.

Of course, $\angle AOC + \angle COB = \pi$. So $\angle APC + \angle CQB = 2 \pi - \frac 1 2 \pi$.


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