Question 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 $AB$.
\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 $AB$.