A Ratio

It is well-known that $P, Q, R$ are collinear ( Simson Line Theorem ).

Moreover, since $∠DPC$ and $∠DQC$ are right angles, the points $D, P, Q, C$ are concyclic and so $∠DCA = ∠DPQ = ∠DPR(∠$ subtende by chord $DQ$ ).

Similarly, since $D, Q, R, A$ are concyclic, we have $∠DAC = ∠DRP.$

Therefore $\Delta DCA∼\Delta DPR$ .

Likewise, $\Delta DAB ∼\Delta DQP$ and $\Delta DBC ∼\Delta DRQ.$ . Then

$\frac{DA}{DC} =\frac{DR}{DP}=\frac{DB\frac{QR}{BC} }{DB\frac{PQ}{BA}} = \frac{QR}{PQ} ·\frac{BA}{BC}$ .

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

So, $\frac{QR}{PQ}=1$

Thus $PQ = QR$