Consider a cyclic quadrilateral $ABCD$. The diagonals $AC$ and $BD$ meet at $P$, and the rays $AD$ and $BC$ meet at $Q$. The internal angle bisector of $\angle BQA$ meets $AC$ at $R$ and the internal angle bisector of $\angle APD$ meets $AD$ at $S$. Prove that $RS$ is parallel to $CD$.
BMO2 2016/17 Geometry
2
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geometry
contest-math
1 Answers
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$$SR \parallel CD \Leftrightarrow \frac {AS}{SD} = \frac{AR}{RC}$$ Since $PS$ is the internal bisector of $\widehat{APD}$, we have $\large\frac {AS}{SD} = \frac {AP}{PD}$. Similarly, $\large\frac {AR}{RC} = \frac {AQ}{QC}$. $$\triangle{QAB} \sim \triangle{QCD} \Rightarrow \frac{AQ}{QC} = \frac{AB}{CD}$$ $$\triangle{PAB} \sim \triangle{PCD} \Rightarrow \frac{AP}{PD} = \frac{AB}{CD}$$ Hence, $SR \parallel CD$.
