MNJ2357 wrote: $\square ABCD$ is a quadrilateral with $\angle A=2\angle C <90^\circ$. $I$ is the incenter of $\triangle BAD$, and the line passing $I$ and perpendicular to $AI$ meets rays $CB$ and $CD$ at $E,F$ respectively. Denote $O$ as the circumcenter of $\triangle CEF$. The line passing $E$ and perpendicular to $OE$ meets ray $OF$ at $Q$, and the line passing $F$ and perpendicular to $OF$ meets ray $OE$ at $P$. Prove that the circle with diameter $PQ$ is tangent to the circumcircle of $\triangle BCD$. Good Problem though identifying the tangency point on paper can be pretty hard to do! Define $Z\equiv \odot(IDF)\cap \odot(IEB)$. We'll show that indeed this is the point of tangency between the two circles . Claim 1: $Z$ lies on the circle $\odot{BDC}$ Proof :\begin{align*}\measuredangle{DZB} &= -(\measuredangle{BZI}+\measuredangle{IZD}) \\ &=-(\measuredangle{BEI}+\measuredangle{IFD}) \\ &=-(\measuredangle{CEF}+\measuredangle{EFC}) \\ &=\measuredangle{FCE}=\measuredangle{DCB} \qquad \square\end{align*} Since $OE=OF$ ; hence Clearly $\triangle{OQE}\cong\triangle{OPF} \implies \odot{PQEF}$ is cyclic trapezium call this circle $\Omega$ Claim 2 : $Z$ lies on $\Omega$ Proof : \begin{align*} \measuredangle{FZE}&= \measuredangle{FZI}+\measuredangle{IZE} \\ &= \measuredangle{FDI}+\measuredangle{IBE} \\ &= \measuredangle{CDI}+\measuredangle{IBC} \\ &= -(\measuredangle{BCD}+\measuredangle{DIB}) \\\ &=-(90^{\circ}+2\measuredangle{BCD}) \\ &= 90^{\circ}-2\measuredangle{BCD}\end{align*} But now note that $\measuredangle{FPE}=90^{\circ}-\measuredangle{EOF}=90^{\circ}-2\measuredangle{BCD} \therefore \measuredangle{FPE}=\measuredangle{FZE}$ which proves the Claim.$\quad \square$ Back to the Original Problem now by Claim 1 and Claim 2 $Z$ lies on both circles, now we'll show that they are tangent.It suffices to show $\measuredangle{ZFE}+\measuredangle{BDZ}=\measuredangle{BZE}=\measuredangle{BIE}$. But since \[\measuredangle{ZFE}+\measuredangle{BDZ}=\measuredangle{ZFI}+\measuredangle{BDZ}=\measuredangle{ZDI}+\measuredangle{BDZ}=\measuredangle{BDI}\] Its well known by Fact-5 that $EF$ is tangent to $\odot{BID}$, hence $\measuredangle{BIE}=\measuredangle{BDI}$ and we are done.$\blacksquare$

Let $E_1 = DI\cap\odot(BCD)$ and $F_1=BI\cap\odot(BCD)$. We prove the following series of claims. Claim: If $EE_1$ meet $FF_1$ at $T$, then $T\in\odot(BDC)$. Proof: Apply the converse of Pascal's theorem on $F_1BDCE_1T$: the intersection points $E,I,F$ are already colinear. $\blacksquare$ Claim: $P,Q,E,F,T$ are concyclic with diameter $PQ$. Proof: Points $P,Q,E,F$ are readily concyclic since $\angle PEQ = \angle PFQ = 90^{\circ}$. To get point $T$, we do a straight angle chasing \begin{align*} \angle ETF &= \angle E_1TF_1 = \angle E_1BF_1 \\ &= \angle BID + \angle IE_1B \\ &= \left(90^{\circ} + \frac{\angle A}{2}\right) + \frac{\angle A}{2} \\ &= 90^{\circ} + \angle A \\[4pt] \angle EPF &= 90^{\circ} - \angle EOF \\ &= 90^{\circ} - 2\angle ECF \\ &= 90^{\circ} - \angle A, \end{align*}which gives $T\in\odot(PQEF)$. $\blacksquare$ Claim: $E_1F_1\parallel EF$. Proof: By Incenter-Excenter lemma, the line through $I$ perpendicular to $AI$ is tangent to $\odot(BID)$. Hence, applying Reim's theorem on $\odot(IIBD)$ and $\odot(BDE_1F_1)$ gives the result. $\blacksquare$ By homothety, these three claims are enough to imply that $\odot(PQ)$ and $\odot(BDC)$ are tangent at $T$.