I think that this problem must be at the High School Olympiad Forum, since it was part of the 2015 Mexican Math Olympiad, and without the use of Ravi's Transformation, this is quiet difficult.

Dear Mathlinkers, if A' is the foot of the A-bissector of ABC, the pencil (D ; A, A', I, E) is harmonic (E being the A-excentre of ABC) ; and we are done... Sincerely Jean-Louis

Math_CYCR wrote: I think that this problem must be at the High School Olympiad Forum, since it was part of the 2015 Mexican Math Olympiad, and without the use of Ravi's Transformation, this is quiet difficult. Indeed it should be, I put it on this forum by mistake

juckter wrote: Indeed it should be, I put it on this forum by mistake I already reported it and Fixed

IsaacJimenez wrote:

Obviously $E$ is the $A$-excenter. Let $R$ be the midpoint of $AD$ and $S$ be the midpoint of $IJ$; we claim that $E, S$ and $R$ are collinear, whence the conclusion follows from reflecting $A$ and $I$ about line $RS$. Put $K$ as the tangency point of the $A$-excircle on $BC$ and $V$ as the antipode of $S$ with respect to the incircle; notice by homothety that $A, S, K$ are collinear. Then $K$ is the center of homothety sending $VS$ to $AD$; hence $K$, $I$ and $R$ are collinear. Now obviously triangles $ARI$ and $EKI$ are similar, from which we get $\frac{RI}{KI}=\frac{AI}{EI}$. Now it suffices to show that this ratio of similitude is equal to the ratio between the inradius and the $A$-exradius, since then triangles $RIS$ and $RKE$ are homothetic, implying the collinearity. But this is clear by the homothety through $A$ that sends the incircle to the $A$-excircle, so we're done$.\:\blacksquare\:$

Proposed by Eduardo Velasco

To clarify jayme's solution... We claim that $(A,A';I,E)=-1$. Performing an inversion about $A$ with radius $\sqrt{AB\cdot AC}$ and reflecting about the $A$ angle-bisector, $A\mapsto P_{\infty}, A' \mapsto M$ and $I\mapsto E$ (where $M$ is the midpoint of $IE$) so that $-1=(P_{\infty},M;E,I)=(A,A';I,E)$ since inversion preserves cross ratio (of course, you can prove this lemma by direct computation, but I think this is the fastest way). Projecting from $D$ onto line $EK$ gives $$(A,A';I,E)\stackrel{D}{=}(P_{\infty},K;DI\cap EK,E)=-1$$which implies that $K$ is the midpoint of $DI\cap EK$ and $E$, so reflecting back over $BC$ gives us what we wanted$.\:\blacksquare\:$

EulerMacaroni wrote: To clarify jayme's solution... We claim that $(A,A';I,E)=-1$. Performing an inversion about $A$ with radius $\sqrt{AB\cdot AC}$ and reflecting about the $A$ angle-bisector, $A\mapsto P_{\infty}, A' \mapsto M$ and $I\mapsto E$ (where $M$ is the midpoint of $IE$) so that $-1=(P_{\infty},M;E,I)=(A,A';I,E)$ since inversion preserves cross ratio (of course, you can prove this lemma by direct computation, but I think this is the fastest way). Projecting from $D$ onto line $EK$ gives $$(A,A';I,E)\stackrel{D}{=}(P_{\infty},K;DI\cap EK,E)=-1$$which implies that $K$ is the midpoint of $DI\cap EK$ and $E$, so reflecting back over $BC$ gives us what we wanted$.\:\blacksquare\:$ Actually the solution is much easier: simply invoke the perpendicularity/angle bisector lemma a few times to prove that $A, I, A', E$ is harmonic and then $AA'$ is the angle bisector of $\angle IDE$.

Let $F$ be the foot of the altitude from $I$ to $BC$. It is well known that $EF$ bisects $AD$ since $E$ is the excenter. But now since $A$, $I$, and $E$ are collinear, $D$, $J$, and $E$ are collinear.

ABCDE wrote: Let $F$ be the foot of the altitude from $I$ to $BC$. It is well known that $EF$ bisects $AD$ since $E$ is the excenter. But now since $A$, $I$, and $E$ are collinear, $D$, $J$, and $E$ are collinear. Isn't that the same as my solution? (Post 11)

Generalization.-Let $I$ be the incenter of an acute-angled triangle $ABC$. Line $AI$ cuts the circumcircle of $BIC$ again at $I_A$. Points $P, Q$ lie on $BC$ such that $AP \parallel IQ$. Let $J$ be the reflection of $I$ through $Q$. Prove that $P, J$ and $I_A$ are collinear.

does it have complex solution?

Math_CYCR wrote: Generalization.-Let $I$ be the incenter of an acute-angled triangle $ABC$. Line $AI$ cuts the circumcircle of $BIC$ again at $I_A$. Points $P, Q$ lie on $BC$ such that $AP \parallel IQ$. Let $J$ be the reflection of $I$ through $Q$. Prove that $P, J$ and $I_A$ are collinear. It is well known that $I_{a}$ is the $A$- excenter of $\triangle ABC$. Let $D$ be the foot of the perpendicular from $A$ to $BC$ and $E$ be the reflection of $I$ over $BC$.Then $\triangle APD$ and $\triangle IRE$ are homothetic triangles.$A,I,I_{a}$ are collinear and from the main problem,$D,E,I_{a}$ are also collinear.So,$I_{a}$ is their homothetic center.Then,$P,J,I_{a}$ are collinear,as desired.

If $P=BC\cap AI$ and $A'$ is the reflection of $A$ across $BC$, then $P$, $J$, $A'$ are collinear. So $D$, $J$ $E$ are collinear if and only if $$\frac{PJ}{JA'}\frac{A'D}{DA}\frac{AE}{EP} =1$$if and only if $$\frac{PI}{IA}\frac{AE}{EP} =1$$if and only if(by angle bisector Theorem and by the fact that $IE$ is a diameter of the second circumference, and so $CE$ is external bisector) $$\frac{AC}{PC}\frac{PC}{AC} =1.$$

Another solution: Note that, by Fact 5, $E$ is the $A$-excenter. Let $M$ be the midpoint of $AD$. Also let $K$ be the $A$-intouch point. Then it is well known that $E,K,M$ are collinear. As $K$ is the midpoint of $IJ$, taking a homothety centered at $E$ that brings $AD$ to $IJ$ immediately gives the result. $\blacksquare$

Let $X$ be $A-$intouch point, Let $IX$ $ \cap$ $ I_AD$ $=$ $I_1$ and $AI$ $ \cap$ $ BC$ $=$ $F$, $T_A$ be the $A-$extouch point, Let $M$ be midpoint of $AD$ and $G$ be the midpoint of $I_AT_A$ $FM$ bisects $IX$ and there exists homothety $\mathcal{H}$ that sends $\Delta DFA$ $ \mapsto$ $ \Delta T_AFI_A$ $\implies$ $\mathcal{H}$ sends $\Delta DFM$ $ \mapsto$ $ \Delta T_AFG$ $\implies$ $M-F-G$ $\implies$ $M-X-I_A$ $\implies$ $IX=XI_1$ $\implies$ $I_1 \equiv I'$ EDIT: just realised, that this is similar to math_pi_rate's solution

Let $F$ $\equiv$ $AI$ $\cap$ $BC$, $P$ $\equiv$ $IJ$ $\cap$ $BC$ Since: ($IEFA$) = $-$ 1, we have: ($IAEF$) = 2 or $\dfrac{\overline{EI}}{\overline{EA}}$ = 2 $\dfrac{\overline{FI}}{\overline{FA}}$ = 2 $\dfrac{\overline{IP}}{\overline{AD}}$ = $\dfrac{\overline{IJ}}{\overline{AD}}$ So: $D$, $J$, $E$ are collinear

solution with @betongblander It is easy to see that $E$ is the $A$-excenter (by incenter-excenter lemma, from $EGMO$), now we supose $AD$ is the height of the triangle - we'll prove that $ED$ touches the perpendicular from $I$ to $BC$ (name it $r$) in $J$. For this, let $M$ be the midpoint of height and let $r$ intersect $BC$ in $G$ and $r$ meet $ED$ in $J'$. It is known that $M, G, E$ are collinear. This follows from Thales theorem that $IG=GJ$, because $M$ is the midpoint of the height and $\dfrac{AM}{MD}=\dfrac{IG}{IJ'}$, hence $J \cong J'$. Note: to prove that $M,G,E$ are collinear you can use homothety with a parallel line to $BC$ trhough $M$. Note 2: It is easier to prove that, if the perpendicular from $E$ to $BC$ is $X$, then $I,X,M$ are collinear, which is analogue to note 1.

Let $ AI \cap BC= K$ Since: $ IJ \parallel AD$, $DK$ bisects $IJ$ therefore $D(AKIJ)=-1=D(AKIE)$ thus $D,J,E$ are collinear