International Mathematical Olympiad

Proof By Simson's Theorem, we know that $P$, $Q$, $R$ are collinear. Moreover, since $\angle DPC$ and $\angle DQC$ are right angles, the points $D$, $P$, $Q$, $C$ are concyclic and so $\angle DCA=\angle DPQ=\angle DPR$. Similarly, since $D$, $Q$, $R$, $A$ are concyclic, we have $\angle DAC=\angle DRP$. Therefore $△DCA\sim △DPR$.



Likewise, $△DAB\sim △DQP$ and $△DBC\sim △DRQ$. Then $$\frac{DA}{DC}=\frac{DR}{DP}=\frac{DB\cdot \frac{QR}{BC}}{DB\cdot \frac{PQ}{BA}}=\frac{QR}{PQ}\cdot \frac{BA}{BC}.$$ Thus $PQ=QR$ if and only if $\frac{DA}{DC}=\frac{BA}{BC}$.

Now the bisectors of the angles $ABC$ and $ADC$ divide $AC$ in the ratios of $\frac{BA}{BC}$ and $\frac{DA}{DC}$ respectively. This completes the proof.