There must be a typo in the question. $Q$ is the circumcircle of traingle $IBC$.

If $Q$ is the circumcenter of $\triangle IBC$, then $Q$ lies on the circumcircle of $\triangle ABC$!

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Absolutely I know that!

Let QP intersect (ABC) at R QR*QP = QI*QI , so angle RIQ = QPI Reflect R wrt IA and get point R' Easy to see that projections of segment R'I on AB , AC are equal to DA , EA , so R'I is perpendicular to PI snd ED PI || ED . done

i haven't drawn all the lines needed in the figure to make it clear $\frac{SM}{MD}=\frac{SI}{IB}=\frac{CS}{CB}=\frac{b}{a+c}$, $SM=AS-AM=\frac{b(c-a)}{2(c+a)}$, hence $MD=\frac{c-a}{2},CD=\frac{b+c-a}{2}$, similarly $BE=\frac{b+c-a}{2}$. By assuming $Q'= AI \cap \odot ABC, R= AO \cap \odot ABC, P'= Q'R \cap BC$ , apply Menelause's theorem to prove $IP' \parallel DE$ . D,E are tangent points with excircles , I don't know if this does any help for reducing the calculation

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By assuming $ Q'= AI\cap\odot ABC, R= AO\cap\odot ABC, P'= Q'R\cap BC $ , apply Menelause's theorem to prove $ IP'\parallel DE $. Would you elaborate on?

Let $X, Y, Z$ be the midpoints of arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$ not containing the vertices of the triangle, respectively. Let $A'$ be the antipode of $A$ w.r.t. $\odot (ABC)$ and denote $R \equiv AC \cap A'Y, S \equiv AB \cap A'Z, P' \equiv XA' \cap BC.$ We will show that $P' \equiv P.$ This will be sufficient, because then the projection of $P$ onto $AI$ will just be $X$ (since $A'X \perp AX$), which clearly lies on the circumcircle. First, it is well-known (Lemma 2) that $D$ and $E$ are just the points of tangency of the $B$-excircle and $C$-excircle with sides $\overline{AC}$ and $\overline{AB}$, respectively. It follows from well-known length relations that $AD = \tfrac{1}{2}(a + b - c)$ and $AE = \tfrac{1}{2}(a - b + c).$ Now, from Pascal's Theorem on cyclic hexagon $CABYA'Z$, it follows that $R, S, I$ are collinear. From Pascal's Theorem on $AXA'YBC$, we have $I, P', R$ are collinear. Therefore, $P' \in RS.$ Hence, we need only show that $RS \parallel DE$, which is equivalent to $AR : AS = AD : AE.$ In order to compute these lengths, we use the Angle-Bisector Theorem: Note that $Y$ is the midpoint of arc $\widehat{AC}$, which implies that $A'R$ bisects $\angle AA'C.$ By the Angle-Bisector Theorem, we obtain the system of equations \[AR + CR = AC, \quad AR : CR = A'A : A'C.\] Solving the system, we find $AR = \frac{AC \cdot A'A}{A'A + A'C}.$ Therefore, \[AR : AS = \frac{AC \cdot A'A}{A'A + A'C} : \frac{AB \cdot A'A}{A'A + A'B} = \frac{AC}{AB} \cdot \frac{2R + A'B}{2R + A'C},\] where $R$ denotes the circumradius of $\triangle ABC.$ Because $\angle A'CB = 90^{\circ} - \angle ACB$, it follows from the Law of Sines that \[AR : AS = \frac{AC}{AB} \cdot \frac{2R + 2R\sin\left(90^{\circ} - C\right)}{2R + 2R\sin\left(90^{\circ} - B\right)} = \frac{AC}{AB} \cdot \frac{1 + \cos C}{1 + \cos B}.\] From the Law of Cosines, we obtain \[AR : AS = \frac{b}{c} \cdot \frac{1 + \frac{a^2 + b^2 - c^2}{2ab}}{1 + \frac{a^2 - b^2 + c^2}{2ac}} = \frac{b}{c} \cdot \frac{\frac{(a + b)^2 - c^2}{2ab}}{\frac{(a + c)^2 - b^2}{2ac}} = \frac{(a + b + c)(a + b - c)}{(a + b + c)(a - b + c)} = AD : AE,\] as desired. $\square$

Dear Mathlinkers, another link http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=149157 Sincerely Jean-Louis

Trivially from homothety we can conclude that B-excircle touches $AC$ at $D$ and C-excricle touches $AB$ at $E$, now $S$ be the A-sharkydevil point and $S'$ be it's extraverted version, now let $I_A$ be the $A$-excenter and let $A'$ be the diametral opposite of $A$ in $(ABC)$, redefine $Q$ as midpoint of arc $BC$ not containing $A$ in $(ABC)$ and $P$ as $A'Q \cap BC$ now we want $IP \parallel DE$, so from I-E Lemma we get that it's the center of $(BICI_A)$. Recall the well known propeties of the Sharkydevil point (such as being a miquel point and lying in $(AI)$, lol). Claim 1: $AS' \parallel DE$ Proof: Project cross ratio from $A$ to get a condition with the midpoint of $DE$, and in general from Ratio Lemma one wants to prove that: \[ \frac{S'B}{CS'}=\frac{SC}{BS}=\frac{p-c}{p-b}=\frac{AD}{EA} \]And then we get our harmonic quadrilateral that projected from $A$ gives desired conclusion. Finish: Note that from invert at $(BIC)$ we get that $A' \to P$ therefore $\angle PIQ=\angle IA'Q=\angle SAQ=\angle S'AQ$ thus getting $IP \parallel AS' \parallel DE$ as desired thus we are done .