(1) First let $\angle ACP = \angle BCQ = \theta$, then $\angle ACQ = \angle BCP = \angle C - \theta$. Also, since $Q_1$ is the reflection of $Q$ with respect to side $C$, we have $\angle QCB = \angle Q_1CB$ and $CQ = CQ_1$. Therefore, $\angle PCQ_1 = \angle PCB + \angle Q_1CB = \angle ACQ + \angle BCQ = \angle C$.

(2) Next let the feet of perpendiculars from $P$ to $BC, CA, AB$ be $M_1, M_2, M_3$ respectively, and let the feet of perpendiculars from $Q$ to $BC, CA, AB$ be $N_1, N_2, N_3$ respectively. Let $S_{\triangle ABC}$ denote the area of $\triangle ABC$. Then,
$$\begin{aligned} S_{\triangle PCQ_1} &= S_{\triangle PCD} + S_{\triangle Q_1CD} \\ \Rightarrow \quad &CP \times CQ \times \sin PCQ_1 = CP \times CD \times \sin PCD + CD \times CQ_1 \times \sin DCQ_1 \\ \Rightarrow \quad &CP \times CQ \times \sin \angle C = CD \times CP \times \sin PCD + CD \times CQ_1 \times \sin DCQ_1 \\ \Rightarrow \quad &CD \times PM_1 + CD \times Q_1N_1 = CD \times (PM_1 + QN_1). \end{aligned}$$
Therefore $CD = \frac{CP \times CQ \times \sin \angle C}{PM_1+QN_1}$.

(3) Similarly it can be shown that $CE = \frac{CP \times CQ \times \sin \angle C}{PM_2+QN_2}$, therefore $\frac{CD}{CE} = \frac{PM_2+QN_2}{PM_1+QN_1}$.

(4) Similarly, $\frac{AE}{AF} = \frac{PM_3+QN_3}{PM_2+QN_2}$ and $\frac{BF}{BD} = \frac{PM_1+QN_1}{PM_3+QN_3}$, therefore:
$$\frac{AE}{EC} \times \frac{CD}{DB} \times \frac{BF}{FA} = \frac{CD}{CE} \times \frac{AE}{AF} \times \frac{BF}{BD} = 1,$$
therefore by Ceva's Theorem, $AD, BE, CF$ are concurrent. Q.E.D.