China Mathematical Competition (Complementary Test)

(1) As shown in Fig. 2, join $NI$, $MI$. Since $PC \parallel MN$ and $P, C, M, N$ are concyclic, $PCMN$ is an isosceles trapezoid. Therefore, $NP = MC$, $PM = NC$.
Join $AM$, $CI$. Then $AM$ intercepts $CI$ at $I$. We have
$$\begin{align*} \angle MIC &= \angle MAC + \angle ACI \\ &= \angle MCB + \angle BCI \\ &= \angle MCI. \end{align*}$$

Fig. 2

Therefore, $MC = MI$. In the same way, $NC = NI$. Then $NP = MI$, $PM = NI$.
This means that $MPNI$ is a parallelogram. Therefore, $S_{\triangle PMT} = S_{\triangle PNT}$, for the two triangles having the same base and height.
On the other hand, $\angle TNP + \angle PMT = 180^{\circ}$, as $P, N, T, M$ are concyclic. Then we have
$$\begin{aligned} S_{\triangle PMT} &= \frac{1}{2} PM \times MT \sin \angle PMT \\ &= S_{\triangle PNT} = \frac{1}{2} PN \times NT \sin \angle PNT \\ &= \frac{1}{2} PN \times NT \sin \angle PMT. \end{aligned}$$
Therefore, $MP \times MT = NP \times NT$.

(2) As shown in Fig. 3, we have
$$\angle NCI_1 = \angle NCA + \angle ACI_1 = \angle NQC + \angle QCI_1 = \angle CI_1N.$$
Therefore, $NC = NI_1$. In the same way, $MC = MI_2$.
From $MP \times MT = NP \times NT$ we get
$$\frac{NT}{MP} = \frac{MT}{NP}.$$
From (1) we know that $MP = NC$, $NP = MC$. Then

Fig. 3
$$\frac{NT}{NI_1} = \frac{MT}{MI_2}.$$
Furthermore,
$$\angle I_1 NT = \angle QNT = \angle QMT = \angle I_2 MT.$$
Therefore, $\triangle I_1 NT \sim \triangle I_2 MT$. Consequently, $\angle NTI_1 = \angle MTI_2$. Then we have
$$\angle I_1 QI_2 = \angle NQM = \angle NTM = \angle I_1 TI_2.$$
This means that $Q, I_1, I_2, T$ are concyclic.