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 6 of 6: Conclude the three lines meet at

O

Previous step Next step

O\in AB^{\perp}\cap \text{bis}(\angle ADP)\cap \text{bis}(\angle PCB)

Detailed analysis

By Step 2, $O$ lies on the perpendicular bisector of $AB$ ; by Step 4, $O$ lies on the internal bisector of $\angle PCB$ ; by Step 5, $O$ lies on the internal bisector of $\angle ADP$ . Hence all three lines pass through the single point $O$ , so they are concurrent.