2003 IMO Problems/Problem 4

Problem

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 $\overline{AC}$ .

Solution

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.

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

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