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 3 of 6: Compute

\angle BOP

and

\angle BCP

in terms of

\beta

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In plain words

The inscribed angle theorem turns the angle $\angle BAP$ at $A$ into a central angle at $O$ , while the angle sum of triangle $BPC$ turns the given angles at $B$ and $P$ into the angle at $C$ .

\angle BOP=2\angle BAP=4\beta,\qquad \angle BCP=180^\circ-(\angle CBP+\angle BPC)=180^\circ-4\beta

Detailed analysis

In the circumcircle of triangle $APB$ , the central angle $\angle BOP$ over chord $BP$ equals twice the inscribed angle $\angle BAP=2\beta$ , so $\angle BOP=4\beta$ . In triangle $BPC$ , the angle sum gives $\angle BCP=180^\circ-\angle CBP-\angle BPC=180^\circ-\beta-3\beta=180^\circ-4\beta$ .