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 2 of 6: Guess the concurrency point: the circumcenter of triangle

APB

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In plain words

The perpendicular bisector of $AB$ automatically passes through the circumcenter of any triangle built on side $AB$ , so this point is the natural candidate to test against the two angle bisectors.

O=\text{circumcenter of }\triangle APB

Detailed analysis

Let $O$ be the circumcenter of triangle $APB$ . Since $OA=OB$ , the point $O$ lies on the perpendicular bisector of segment $AB$ by definition; it remains to show $O$ also lies on the internal bisectors of $\angle ADP$ and $\angle PCB$ .