Problem 1

Consider the convex quadrilateral

ABCD

. The point

P

is in the interior of

ABCD

. The following ratio equalities hold:

\angle PAD : \angle PBA : \angle DPA = 1 : 2 : 3 = \angle CBP : \angle BAP : \angle BPC

. Prove that the following three lines meet in a point: the internal bisectors of angles

\angle ADP

and

\angle PCB

, and the perpendicular bisector of segment

AB

Step 4 of 6: Deduce that

B,C,P,O

are concyclic and that

CO

bisects

\angle PCB

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In plain words

Opposite angles summing to $180^\circ$ is exactly the cyclic-quadrilateral criterion, and equal radii $OB=OP$ subtend equal angles from any point on that circle.

\angle BOP+\angle BCP=180^\circ\implies B,C,P,O\text{ concyclic}\implies \angle BCO=\angle OCP

Detailed analysis

Since $\angle BOP+\angle BCP=4\beta+(180^\circ-4\beta)=180^\circ$ , quadrilateral $BCPO$ is cyclic. As $OB=OP$ are both radii of the circumcircle of $APB$ , the equal chords $OB,OP$ subtend equal angles from $C$ on circle $BCPO$ , so $\angle BCO=\angle OCP$ ; hence $CO$ bisects $\angle PCB$ .