MathLabs

Problem 1

Consider the convex quadrilateral ABCDABCD. The point PP is in the interior of ABCDABCD. The following ratio equalities hold: ∠PAD:∠PBA:∠DPA=1:2:3=∠CBP:∠BAP:∠BPC\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 ∠ADP\angle ADP and ∠PCB\angle PCB, and the perpendicular bisector of segment ABAB.
Step 1 of 6: Name the angles from the two given ratios
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In plain words

Each ratio 1:2:31:2:3 pins down a single unknown, turning the hypothesis into two triangles whose angles are all multiples of one parameter.

α=∠PAD, 2α=∠PBA, 3α=∠DPA;β=∠CBP, 2β=∠BAP, 3β=∠BPC\alpha=\angle PAD,\ 2\alpha=\angle PBA,\ 3\alpha=\angle DPA;\qquad \beta=\angle CBP,\ 2\beta=\angle BAP,\ 3\beta=\angle BPC
Detailed analysis

From the first ratio set ∠PAD=α\angle PAD=\alpha, ∠PBA=2α\angle PBA=2\alpha, ∠DPA=3α\angle DPA=3\alpha; from the second set ∠CBP=β\angle CBP=\beta, ∠BAP=2β\angle BAP=2\beta, ∠BPC=3β\angle BPC=3\beta, for some α,β>0\alpha,\beta>0.