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 5 of 6: Repeat the argument on the

D

side:

DO

bisects

\angle ADP

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In plain words

Swapping the roles of $A\leftrightarrow B$ and $D\leftrightarrow C$ , and $\beta$ for $\alpha$ , repeats exactly the same computation.

\angle AOP=2\angle ABP=4\alpha,\qquad \angle ADP=180^\circ-(\angle PAD+\angle DPA)=180^\circ-4\alpha\implies A,D,P,O\text{ concyclic}\implies \angle ADO=\angle ODP

Detailed analysis

By the same reasoning with $A,D,\alpha$ in place of $B,C,\beta$ : the central angle $\angle AOP=2\angle ABP=4\alpha$ , and the triangle angle sum gives $\angle ADP=180^\circ-\angle PAD-\angle DPA=180^\circ-\alpha-3\alpha=180^\circ-4\alpha$ . Their sum is $180^\circ$ , so $A,D,P,O$ are concyclic, and $OA=OP$ gives $\angle ADO=\angle ODP$ , so $DO$ bisects $\angle ADP$ .