IRL ABooklet

Geometry Difficulty 5.6 AIME, harder Prove it Ireland

A tangent is drawn at $A$ to a circle centre $O$ . Point $D$ is on this tangent and $B$ , $C$ are on the circle such that $B$ , $C$ and $D$ are collinear. Points $P$ and $Q$ are the circumcentres of triangles $ACD$ and $ABD$ , respectively.
Prove $\angle AOB = \angle APD = \angle BQD$ .

Solution

Because $P$ is the circumcentre of triangle $ACD$ , we have
$\frac{1}{2} \angle APD = \angle ACD = 180^\circ - \angle ACB.$
Because $O$ is the circumcentre of triangle $ABC$ we see that
$\frac{1}{2} \angle AOB = 180^\circ - \angle ACB$
and we obtain $\angle APD = \angle AOB$ .

Because $Q$ is the circumcentre of triangle $ABD$ , we have $\frac{1}{2}\angle BQD = \angle BAD$ . From the alternate segment theorem we know that $\angle BAD$ is equal to an inscribed angle that stands on the arc $BA$ that contains $C$ , which is equal to one half of the central angle $\angle AOB$ . This shows that $\angle BQD = \angle AOB$ . The reasoning in other configurations is slightly different, but similar.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.