2003 IMO Problems/Problem 4

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Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, and $R$ be the feet of perpendiculars from $D$ to lines $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Show that $PQ=QR$ if and only if the bisectors of angles $ABC$ and $ADC$ meet on segment $\overline{AC}$.


Clearly $PQR$ is the Simson Line and $APDQ$, $BPDR$, $CQDR$ is cyclic. By angle chasing we have $\triangle DPQ\sim\triangle DBC$, $\triangle DQR\sim\triangle DAB$. Then by $PQ=QR$ we have $\frac{DC}{CB}=\frac{DQ}{QP}=\frac{DQ}{QR}=\frac{DA}{AB}$. Rearranging and using the angle bisector theorem we are done.

See Also

2003 IMO (Problems) • Resources
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
All IMO Problems and Solutions