Very similiar to the USA december TST. Infact it is just a easier version of this problem!

Let G be the intersection point of $AI$ and $BC$. $ \angle CAG = \angle BAG = 90 - (\dfrac{\angle B + \angle C }{2}) $ $ \angle DFC = 90 - \dfrac{\angle C}{2} $. $ \angle DPG = \angle APF = \angle DFC - \angle PAF = \dfrac{\angle B}{2}$ and similarly $\angle DQG = \dfrac{\angle C}{2}$. Draw $BI$, $CI$ and $ID$. We can easily see that $\angle IDP = \dfrac{\angle C}{2} $. That means the line $ID$ is tangent to the circumcircle of triangle $PDQ$ at $D$. We also know that $ BC \perp ID $ and that means the circumcenter of triangle $DPQ$ lies on $BC$. Lets call it $M$. By some angle chasing, we can see that $IPDB$ and $QDIC$ are cyclic and segments $IB$ and $IC$ are their circumcircles' diameter respectively. Hence $\angle IPB = \angle IQC = 90^\circ $. Let the point $L$ be the midpoint of segment $IQ$. Then $ML \perp IQ $ so we see that $PB // ML // QC$. Let $ML $ intersects $BQ$ at the point $X$. $L$ is the midpoint of $IQ$ so $X$ must be the midpoint of segment $BQ$. Finally look at the triangle $BQC$. $MX$ // $QC$ and X is the midpoint of $BQ$ hence $M$ is the midpoint of $BC$ and we are done.

put M is the midpoint of BC easy to prove AIB similiar AEP so BIPD is incribed => angle BPA = 90 similar AFQ similar AIC => angle FQA = 1/2angle ACB put S is the midpoint of AB. we can prove SM is the middle line of triangle ABC => angle MSB = angle BAC S is the midpoint of AB and angle BPA = 90 => ASP is isosceles => S,P,M are on a line=> PM//AC => PMD= 2PQD. so the circumcenter of DPQ triangle is the midpoint of BC.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/91.%20Cercles.pdf p. 3-5. Sincerely Jean-Louis

Lemma-1: $I, B, D, P$ and $I, C, Q, D$ concyclic. Proof: We know that $\measuredangle IBD = \frac{\hat{B}}{2}$ and in $\Delta APE$, $\measuredangle PAE = \frac{\hat{A}}{2}$ and$\measuredangle PEA = 90 + \frac{\hat{C}}{2}$. Then, $\measuredangle IPE = \frac{\hat{B}}{2}$. SO $I, B, D, P$ are concyclic. Same thing goes for $I, C, Q, D$ as well. From Lemma-1 we see that $BP \perp AP$ and $CQ \perp AP$ so $BP \parallel CQ$. Le $R=PM \cap CQ$ where $M$ is the midpoint of $BC$. It's easy to see that $\Delta BPM \cong \Delta CRM$, so $PM=MR=QM$. Also, $$\measuredangle DPQ = \frac{\hat{B}}{2} = 90- \measuredangle QDM = 90- \frac{\hat{B}}{2}$$and $$\measuredangle DQP = 90- \measuredangle PDM$$similarly. Lemma-2 In a triangle $ABC$, a point $O$ satisfies $OA=OC$, $\hat{A} = 90 - \measuredangle OBC$ and $\hat{C} = 90 - \measuredangle OBA$, then $O$ is the circumcenter of $\Delta ABC$. Proof: Let those equalities hold and for $H \in AC$, let $BH \perp AC$. then $\measuredangle HBA = \measuredangle OBC$. It's seen that $BH$ and $BO$ are isogonal conjugates and $BH$ is altitude, then circumcenter of $\Delta ABC$ must be on $BO$. From this, $O$ must be the circumcenter. From Lemma-2, we easily see that $M$ is the circumcenter of $\Delta DPQ$

Let the perpendicular to $DE$ through $P$ meet $CD$ at $K$, and let $DE$ intersect $CI$ at $J$. I will aim to show $BD=CK$. Since $PK \perp DE$ and $CI \perp DE$, $PK \parallel CI$. Then, $\frac{CK}{CD} = \frac{JP}{JD}$ (*). Also, $\angle JIP = \angle CAI + \angle ACI = 90 - \frac{\angle ABC}{2} = \angle BID$, and $\angle IDB=90$. Thus, $\triangle JIP \sim \triangle DIB$, so $\frac{JP}{IJ} = \frac{BD}{ID}$. Substituting for $JP$ into (*), we have $CK = BD \cdot \frac{IJ}{ID} \cdot \frac{CD}{JD} = BD$ since $\triangle IJD \sim \triangle CJD$. Thus, $CK=BD$, so the midpoint $M$ of $BC$ is also the midpoint of $DK$. But since $\angle DPK=90$, this implies that $MP=MD$. Similarly, $MQ=MD$. Thus, $M$ is the circumcenter of $\triangle DPQ$.