I've managed to reduce it to showing that AT and AL are perpendicular: K is the intersection of B1C1 and PQ, so the polar of P with respect to the incircle of ABC passes through the poles of B1C1 and PQ, which are A and T respectively, so the polar of K with respect to the incircle is AT, but it's well-known that the polar of a point with respect to a circle is perpendicular to the line joining the point to the center of the circle, so AT perpendicular to IK, so in order to show that IK || AL "all" () we need to show is that AT is perpendicular to AL as well. This might not be significantly easier, but it's still progress, isn't it? .

I'm sort of stuck right now, but here are some things I observed (I drew a dynamic sketch with Euklides): Let T' be the analogous of T (i.e. let A₁B₁1 cut MN at T'). Let MN cut BC at S. Then the following hold: (1) ATA₁T' is cyclic; (2) <SAT = <TAB; If we manage to prove these, then I think it's kind of done.

I think I can prove (1) from the previous post: Let the line through T parallel to AB cut AA₁ at R and let the line throguh T' parallel to AC cut AA₁ at R'. N is the midpoint of AC₁ and TR and AB are parallel, so the lines TR, TA, TN, TC₁ form a harmonic cross-ratio, so the points R, A, X, A₁ also form a harmonic cross-ratio (X is the intersection of AA₁ and TT'). We prove the same for R', A, X, A₁ and we get R=R', so A₁TRT' is obtained from A₁C₁AB₁ by a homothety, so RT and RT' are tangent to the circumcircle of A₁TT', so A₁R is a symmedian in the triangle A₁TT', and from the fact that A is the harmonic conjugate of A₁ with respect to (R, X), we easily find that A is on the circle A₁TT'.

I guess the other part isn't tht hard either (part (2) from 2 posts ago): First of all, the tangents to the circumcircle of ATT' through T and T' and the diagonal AA₁ are concurrent. This is called a hramonic cyclic quadrilateral and it follows that the tangents to its circumcircle through A and A₁ and its diagonal TT' are also concurrent, so SA is tangent to the circumcircle, so <SAT=<AT'T=<TAB (this is true because: <TNA=\pi-<ANM=\pi/2+A/2=<TAT'=><TAB=<TAN=<AT'T).

My friends and I have the proof below: After the arguments Grobber just states. we know that,AT=QT=PT(TQ,TP are tangents to O) Draw a tnagent line to circle T at point A,intersect PQ at S.Let A1C1 interesect circle T at U(outside the circle O). Because circle T and O are orthogonal, Consider the inversion with point Aand radius AB1,under which the circle O does not move.As we can see,the circle T changes to a line passes through O.And U' the inversion of U lies on the circumcircle of AB1C1,so AU'O=AC1O=90, since the angle doesnt change under the inversion,AU must be the diameter of circle T. Next, we draw a circle with center S and radius AS,therefore the circle T is orthogonal to the circles O and S.And the radical axis of O and T pass through the center of circle S.It can be proved that,(this is not difficult so i just dont type them),circle S is also orthogonal to circle O. So,if we know already the circle S is orthogonal to circleO.By the same argument above,if circle S intersect B1C1 at V,we can claim that S is the midpoint of AV. By now,we conclude that,ST is parallel to UV,then ST bisects segments AC1 and AB1,the complete the proof. This problem is just so cool that we thought about it all afternoon！ Best regards, chao

Consider $A$ as a degenerate circle, then we will have $MN$ is the radical axis of $A$ and $\odot I$. So we know that $T$ is the circumcenter of $\triangle APE$, furthermore we have $L$ the radical center of $A, \odot I$ and $\odot T$. Therefore $AL$ is tangent to $\odot T$ at $A$, which means $AL \perp AT$. On the other hand, it is not hard to see that $AT$ is the polar of $K$ with respect to $\odot I$, hence $IK \perp AT$. Finally, we have $AL \parallel IK$.

Dear Mathlinkers, http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1137818265&t=161416 Sincerely Jean-Louis

Note that $B_1C_1$ is the polar of $A$ on the incircle, and $PQ$ is the polar of $T$. Thus, by La Hire, $TA$ is the polar of $K$. Let $IK$ meet $TA$ at $S$, so $IS\perp TA$. Now, the idea is that $NM$ is the radical axis of $(A)$ and the incircle. Thus, since $T$ and $L$ lie on said radical axis, we have $$TA=TQ=TP$$so $T$ is the circumcenter of $AQP$. Furthermore, we have $LA^2=LQ\cdot LP$. Thus, if we let $\angle LAQ=\theta$, we have $=\angle LPA=\theta$ as well. However, since $T$ is the circumcenter of $AQP$, we have $\angle ATQ=2\angle APQ=2\theta$. Finally, $TA=TQ$, so $\angle TAQ=90-\theta$, so $\angle TAL=90-\theta+\theta=90$. Thus, $TA\perp AL$, and since $TA$ is the polar of $K$ on the incircle, $TA\perp IK$, and we are done.