interesting to see Mixtilinear at Junior 1)$M,D,I,E$ lie on a circle 2)Sinse $DE//PQ$ we have $M,K,I$ collinear where $K$ is the midpoint of $DE$

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Let $IM \cap DE=K,IM \cap BC=L$ and $C=2\theta,B=2\beta$. $\beta=\angle IBC=\angle ABI=\angle PMI \implies B,D,L,M$ are cyclic. $\theta=\angle ICB=\angle ACI= \angle QMI \implies C,E,L,M$ are cyclic. $\beta+\theta=\angle AQI=\theta+\angle QIC \implies \angle QIC=\beta$ $\beta+\theta=\angle API=\beta+\angle PIB \implies \angle PIB=\theta$ $\angle ICB=\theta=\angle PIB=\angle PMB=\angle DLB \implies DL\parallel IC$ $\angle IBC=\beta=\angle QIC=\angle QMC=\angle ELC\implies EL\parallel IB$ So $IDLE$ is parallelogram and $IL \cap DE=K$ $\implies IK=KL$ and $DK=KE$.

This one was easy imo Draw the lines $BI,CI$ and $AI$. Let $\angle {PBI}= \alpha , \angle {ICQ} = \beta$ . This implies that $\alpha = \angle{PMI} , \angle{IMQ}= \beta$ . And since $P,I,Q$ are collinear and $AP=AQ$ , $I$ is the midpoint of $PQ$ and $\angle{APQ}= \angle {AQP}= \alpha+ \beta$ $\implies \angle{PIB}=\alpha , \angle{QIC}= \beta$ . So $\angle{DIC}= \angle{180- \alpha- \beta} , \angle{PMQ}= \alpha+ \beta$ $\implies IDME$ cyclic. This implies $PQ\parallel DE$, therefore $MI$ bisects $DE$ . The last part is obvious if you look at the triangle $PMQ$.

The condition implies that $P,Q$ are the $A $mixtilinear touch points with the sides and $M$ is the touchpoint of the mixtilinear circle with the circumcircle. Trivial angle chasing yields $DE \parallel PQ$ but since $I$ is midpoint of $PQ$, the midpoint of $DE$ must lie on this due to a property of trapeziums.