$\angle CNQ=\angle RPQ=180^o-(90^o-\frac{1}{2}\angle A)-(90^o-\frac{1}{2}\angle B)=90^o-\angle C \Rightarrow CN=CQ$,and obviously,$AP=AR$;$BM=BK$ Let $\odot (MNK) \cap AC=N$and$R$;$D$ is the midpoint of $PM$;$AM=AN=k$ $\therefore MB=MK=a-k;CN=CQ=b-k$ $\Rightarrow QK=PM=NR=a-(c-k)-(b-k)=a-b-c+2k$ $\Rightarrow AD=AP+\frac{PM}{2}=\frac{b+c-a}{2}$ It means point $P$ is fixed. Then we prove $C,T,D$ are collinear. $\because \triangle PMT\sim \triangle NQT$ $\therefore \frac{TN}{TP}=\frac{NQ}{PM} \Rightarrow TN=\frac{TP\cdot NQ}{PM}$ $\therefore \frac{MT}{TN}\cdot \frac{NC}{CA}\cdot \frac{AD}{DM}$ $=\frac{TM}{TP}\cdot \frac{PM}{NQ}\cdot \frac{NC}{CA}\cdot \frac{AD}{DM}$ $=\frac{sin(90^o-\frac{B}{2})}{sin(90^o-\frac{A}{2})}\cdot \frac{2DM}{CN\cdot sin\frac{C}{2}}\cdot \frac{NC}{CA}\cdot \frac{AD}{DM}$ $=\frac{cos\frac{B}{2}}{cos\frac{A}{2}\cdot sin\frac{C}{2}}\cdot \frac{AD}{AC}$ $=\frac{\sqrt{\frac{(a+c)^2-b^2}{4ac}}}{\sqrt{\frac{(b+c)^2-a^2}{4bc}}\cdot \sqrt{\frac{c^2-(a-b)^2}{4ab}}}\cdot \frac{b+c-a}{2b}$ $=\sqrt{\frac{4b^2[(a+c)^2-b^2]}{[(b+c)^2-a^2][c^2-(a-b)^2]}}\cdot \frac{b+c-a}{2b}$ $=\sqrt{\frac{(a+b+c)(a-b+c)4b^2}{(a+b+c)(b+c-a)(b+c-a)(a-b+c)}}\cdot \frac{b+c-a}{2b}$ $=\frac{2b}{b+c-a}\cdot \frac{b+c-a}{2b}$ $=1$ $\therefore C,T,D$ are collinear, so we are done.

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Another approach:(easier) $\angle ANM=\angle AMT\Rightarrow sin\angle CNT=sin\angle TMP$ The same,$sin\angle CQT=sin\angle TPM$ And in $\triangle TMD$ we have$\frac{DM}{sin\angle DTM}=\frac{DT}{sin\angle TMD}$ $\Rightarrow DM\cdot sin\angle TMD=DT\cdot \angle DTM$ Also, in $\triangle SNT$ we have$CN\cdot sin\angle CNT(sin\angle DMT)=CT\cdot sin\angle CTN(sin\angle DTM)$ So $\frac{DM}{CN}=\frac{DT}{CT}$ The same,$\frac{DP}{CQ}=\frac{DT}{CT}$ $\therefore PD=MD$

Let $RP \cap KM=X$,obviously,$TPXM$ is parallelogram, and we apply Pascal's to $NRPMKQ$ $\Rightarrow CTX$ are collinear, so we have done