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 $\triangle DCA \sim \triangle DPR$.

Likewise, $\triangle DAB \sim \triangle DQP$ and $\triangle DBC \sim \triangle DRQ$. 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.