# 2007 JBMO Problems/Problem 2

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem 2

Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$.

## Solution

Let I be the intersection between $(DP)$ and the angle bisector of $\angle{DAP}$ So $\angle{CAI}=\angle{PAI}=36/2°=18°$ So $\angle{CAI}=18°=\angle{CBD}=\angle{CBI}$ We can conclude that $A,B,C,I$ are on a same circle. So $\angle{ICB}=180-\angle{IAB}=180-\angle{IAC}-\angle{CAB}=180-18-72=90$ Because $\angle{CBD}=18$ and $\angle{CDB}=36$ we have $126=\angle{DCB}=\angle{ICB}+\angle{ICD}=90+\angle{ICD}$ So $\angle{ICD}=36=\angle{BDC}=\angle{IDC}$ So $I$ is on the angle bisector of $\angle{DAP}$ and on the mediator of $DC$. The first posibility is that $I$ is the south pole of $A$ so $I$ is on the circle of $DAC$ but we can easily seen that's not possible The second possibility is that $DAC$ is isosceles in $A$. So because $\angle{DAC}=36$ and $DAC$ is isosceles in $A$ we have $\angle{ADC}=72$. So $\angle{APD}=180-\angle{PAD}-\angle{PDA}=180-36-(\angle{ADC}-\angle{PDC})=180-36-72+36=108$