Let $A'$ denote the reflection of $A$ in $I$. $Q$ lies on the polar of $A' \implies$ the polar of $Q$ is precisely $DA'$, where $D$ is the tangency point of $(I)$ and $BC$, i.e. the polar of $Q$ is the reflection of $AD' \equiv ANa$ on $I$, where $Na$ is the Nagel point of $ABC$. But $I$ is the Nagel point of the medial triangle, i.e. $IM \parallel ANa \perp IQ.$

Yep. You can also phrase this without Nagel points: let $P$ be the point on the incircle so that the tangent at $P$ is parallel to $BC$. Let $Q' = PP \cap EF$. The poles of these two lines are $A$ and $P$ meaning $Q'$ is the pole of $AP$. But $AP \parallel IM$ (just extend $AP$ to $BC$) so we're done. This is just reflecting $Q$ rather than $A$ which I think is slightly more motivated. It's also straightforward to complex bash with the incircle as the unit circle.

we can also do this : let $D'$ be the reflect point $D$ about $I$ and let $AD'$ intersect the incircle at $T$ now we have $ (D'TEF)=-1$ now if we reflect these four points about $I$ we will have $ (DT'GH)=-1 $ so we have that $QT'$ is tangent to the incircle where $T'$ is the reflect of $T$ about $I$ and now we have that $ DT' \perp QI , DT' \parallel D'T \parallel MI $ so we are done

We can also use Barycentrics, taking $ABC$ as reference triangle and we get that $Q$ = $(0,t,1-t)$ where $t$ equals $[(s-b)(s-2c)/(s-c)(s-2b) -1]^{-1}$. Now applying EFFT or redefining points and using $ID^2$ = $DM.DQ$ we are done.

Too easy for a TST 3rd I give a solution using Cartesian coordinates. Let $I=(0,0)$ and let the incircle be a unit circle. let $D = (-1,0)$, $E=(\cos x, \sin x)$ and $F=(\cos y, \sin y)$ By straightforward computations, $G = (-\cos x,- \sin x)$, $H = (-\cos y, -\sin y)$, $B = (\frac{1+\sin y}{\cos y},-1)$, $C=(\frac{1+\sin x}{\cos x},-1)$ $\implies M=(\frac{\frac{1+\sin x}{\cos x}+\frac{1+\sin y}{\cos y}}{2},-1)$ and $Q = \frac{\cos y - \cos x + \cos x \sin y - \cos y sin y}{\sin x - \sin y}$ Now by straightforward calculation, $IM \perp IQ$

Dear Mathlinkers, a synthetic proof by considering only the Pascal's theorem three times is possible.... Sincerely Jean-Louis

Let $X=$point of contact of incircle with $BC$, $X`=$Antipode of $X$ wrt incircle , $M=$Midpoint of $AI$, $M`=$Midpoint of $IX`$ , $T= AX` \cap BC$. Invert with center $I$ and radius $r$ equal to inradius of $\triangle ABC$. Let $\overleftrightarrow{GH}$ map to $\Omega$ and $\overleftrightarrow{BC}$ map to $\Omega`$ .Consider a homothety of ratio $(-1)$ $\mathcal{H}$ sending $\Omega$ to $\mathcal{H} (\Omega)$ and $\Omega`$ to $\mathcal{H} (\Omega`)$ .Note that $\mathcal{H} (\Omega)$ is a circle with diameter $IA$ and $\mathcal{H} (\Omega`)$ is a circle with diameter $IX`$ .Let $Q`$ denote the image of $Q$ after the composition of inversion and homothety. Notice that $(I,Q`)$ $ \in \mathcal{H} (\Omega`)\cap \mathcal{H} (\Omega)$.So, it suffices to show that $IQ` \perp IM$. But , $\overleftrightarrow{IQ`}$ is the radical axis of $\mathcal{H} (\Omega`)$ and $\mathcal{H} (\Omega)$ .So, it suffices to show that $MM` || IM$. This is same as showing $AX` || IM$, ,ie; $AT || IM$ .But this is obvious from the fact that $T,X$ are isotomic conjugates wrt $\overline{BC}$. Hence , we are done.

let $A_{1}$ be the intersection of tangents from $G,H$ wrt the incircle.Obviously $Q$ is on the polar of $A_{1}$ so by La Hire s theorem we have that $A_{1}$ is on the polar of $Q$,thus all we need to prove that $A_{1},I,M$ are collinear.To prove this extend the before mentioned tangents to cut the sides $AC,AB,BC$,$S,R,E,F$ (the last two intersections are on the side $BC$)respectively. $A_{1}SAR$ is a rhombus thus $A_{1}I$ is the angle bisector of $\angle RA_{1}S$ so we need to prove that $A_{1}M$ is the angle bisector in $\triangle FEA_{1}$ which is not hard (length chasing by parallel sides)

v_Enhance wrote: It's also straightforward to complex bash with the incircle as the unit circle. Indeed, for me it was an excellent complex number tutorial! [asy][asy] unitsize(3cm); pointpen=pathpen=black; pointfontpen=fontsize(10pt); pair A=D("A",dir(110),NW), B=D("B",dir(200),SW), C=D("C",dir(-20),SE), I=D("I",incenter(A,B,C),NE); path uc=D(incircle(A,B,C)); pair D=D("D",IP(uc,D(B--C))), E=D("E",IP(uc,D(C--A)),NE), F=D("F",IP(uc,D(A--B)),NW), G=D("G",2*I-E), H=D("H",2*I-F), Q=D("Q",extension(B,C,G,H),W), M=D("M",(B+C)/2); D(B--Q--I--M); [/asy][/asy] Let the incircle, which we set as the unitcircle, touch $\overline{BC}$ at $D$. Then $g=-e,h=-f$ and $b=\frac{2df}{d+f},c=\frac{2de}{d+e}$, so $$m=\tfrac{1}{2}(b+c)=\frac{d(de+df+2ef)}{(d+e)(d+f)}.$$Now use the chord intersection formula to find $$q=\frac{2dgh-d^2(g+h)}{gh-d^2}=\frac{d(de+df+2ef)}{ef-d^2},$$so $\frac{m}{q}=\frac{ef-d^2}{(d+e)(d+f)}$ and $$\frac{\overline{m}}{\overline{q}}=\frac{\frac{1}{ef}-\frac{1}{d^2}}{\left(\frac{1}{d}+\frac{1}{e}\right)\left(\frac{1}{d}+\frac{1}{f}\right)}=\frac{d^2-ef}{(d+e)(d+f)}=-\frac{m}{q},$$so $\frac{m}{q}\in i\mathbb{R}$, and thus $IQ\perp IM$.

COMPLEX BASH! Pretty straightforward Consider $I$ as origin and the Incircle as unit Circle. Nice Exercise for a beginner like me

Dear Mathlinkers, also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=598559 Sincerely Jean-Louis

Let \(N\) be the midpoint of \(GH \Rightarrow DNIQ\) is cyclic \(\Rightarrow \angle DIQ = \angle DNG\) -(i) Also, angle chasing tells that: \(\Delta BIC \sim \Delta DGH \Rightarrow \angle IMB = \angle DNG\) -(ii) \(\therefore\) from (i, ii) we have: \(\angle DIQ = \angle IMB\), done.

Let $CI$ and $BI$ meet $EF$ at $U$ and $V$, it is well-known that $BVUC$ lies on a circle with center $M$. Let $QI$ meet $EF$ at $P$, because $FEHG$ is a rectangle we have that $QI\cong IP$, and we are done by the converse of Butterfly.

skipping a lot of details and some directed stuff but main idea is here: Let $N$ be the midpoint of arc $BC$ excluding $A$ and $S$ its antipode. Let $SI$ intersect the circumcircle of $ABC$ at $T$ - note that $T$ is the $A$-mixtilinear touchpoint. Now, let $NT$ intersect $BC$ at $K$. By radical axis on the circumcircle of $ABC$, the circle with diameter $NI$, and the $A$ fact $5$ circle, we have that $IK$ is perpendicular to $AN$. Now, we claim that the triangles $IKQ$ and $INM$ are spirally similar - this is sufficient. Note that since $\angle{KIN} = \angle{KMN} = 90^\circ$, we have that $KIMN$ is cyclic and thus $\angle{IKM} = \angle{INM}$. Thus, it is sufficient to prove that $\dfrac{IK}{IN} = \dfrac{KQ}{MN}$. Note that $\dfrac{IK}{IN} = \dfrac{TI}{TN} = \dfrac{AI}{AS}$. Now, let $\angle{IKM} = a$ and $\angle{\frac{A}{2}} = b$. Note that: \begin{align*} XK &= IY \csc{a} \\ &= IA \csc{a} \sin^2{b} \end{align*}\begin{align*} MN &= NS \sin^2{b} \\ &= AS \csc{a} \sin^2{b} \end{align*}Thus, $\dfrac{XK}{MN} = \dfrac{AI}{AS}$ and we are done.

Denote the intersection of the tangents from $G,H$ as $A'$, the antipode of $D$ as $D'$, and the A-extouch point as $X$. It is well known that $AD'X$ is a line, and since $IM$ is the midline of triangle $DD'X$, $AD'\parallel IM$. Note that we wish to show $IQ\perp IM$, or $IM$ is parallel to the polar of $Q$. However, as $Q$ lies on the polars of $A'$ and $D$, its polar is $A'D$. So, we wish to show $IM\parallel A'D$. Now, reflect over the incenter. Line $IM$ is preserved, and $A'D$ goes to $AD'$, so this is equivalent to $AD'\parallel IM$, which we already know.

Easy trig bash... Let WLOG $AC > AB$ i.e. $b > c$. Let $A,B,C$ denote the angles of $\triangle ABC$ and let $a,b,c$ denote the sides of $\triangle ABC$. Let $R$ and $r$ denote the circumradius and inradius of $\triangle ABC$ respectively. $\therefore MD = \frac{b-c}{2} = R(sinB - sinC) = 2Rsin\left(\frac{B-C}{2}\right)sin\frac{A}{2}$ Now we'll compute $DQ$ Consider $\triangle HDQ$. It follows easily by angle chasing that $\angle QHD = \frac{C}{2}$ and $\angle HQD = \frac{B-C}{2}$. Also $HD = 2rsin\frac{B}{2}$. Now by sine rule in $\triangle HQD$ we obtain $\frac{HD}{sin\frac{B-C}{2}} = \frac{QD}{sin\frac{C}{2}}$ $\therefore QD = \frac{2rsin\frac{B}{2}sin\frac{C}{2}}{sin\left(\frac{B-C}{2}\right)}$ $\therefore QD.DM = 4Rrsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} = r^2 = ID^2 $. Also $ID \perp QM$. $\therefore \angle MIQ = 90$ as desired.

Suppose the incircle is tangent to $BC$ at $D$, and let $K\neq D$ be another point on the incircle such that $QK$ is tangent to the incircle. Then $GKHD$ is a harmonic quadrilateral. Let $D'$ and $K'$ be the reflections of $D$ and $K$ about $I$, then projecting through $I$, we can see $ED'FK'$ is a harmonic quadrilateral, so $AD'K'$ are collinear. Since $AD'$ also passes through the extouch point $L$ on $BC$, we have $KD\parallel IM\parallel D'L$ because $M$ is the midpoint of $DL$. Therefore $IQ\perp IM$.

Let $D'$ be the reflection of $D$ across $I$ and let $AD'$ meet the incircle again at $K$ and $BC$ at $X$. Then, we see that $D'EKF$ is harmonic. If $K'$ is the reflection of $K$ across $I$, then $DGHK'$ is also harmonic. This means that the intersection of the tangents from $K'$ and $D$ is on $GH$. Thus, $Q$ is the pole of $FD$, so $IQ \perp FD$. We also know $FD || D'K$ and $D'K || IM$, since $IM$ is the midline in $D'DX$. Therefore, $IQ \perp IM$.

Reflect $Q$ over $I$ to get $Q'$, and let $P$ be the antipode of $D$; it suffices to prove $IQ'\perp IM$. But note $Q', A$ are polar so $AP$ is the polar of $Q'$, thus $IQ'\perp AP$ and we want to prove $AP||IM$. Letting $K=AP\cap BC$, we have that $K$ is the excircle tangency point, so $BD=CK$; thus, $DM=MK$ and $PI=ID$, so $IM||PK$, and we're done.

Let $\triangle DEF$ be the contact triangle. Reflect $A$ over $I$ to get $A'$ which has line $GH$ as it's polar. And so $Q$ lies on the polar of $A'$ and by La Hire, $A'$ lies on the polar of $Q$. Moreover, $D$ being the tangency point of line $BC$ with $\odot(I)\implies D$ lies on the polar of $Q$ too and thus line $A'D$ is the polar of $Q$ and so we have $IQ\perp A'D$. Now $D'$ be the $A-$extouch point and $X=ID\cap\odot(I)$. And as $A'$ is the reflection of $A$ over $I$, we have $DA'\parallel AX\equiv AD'\parallel IM$ where the last parallelism (is this even a valid word?) follows from the Diameter of the Incircle Lemma and we are done.

Let $D$ be on the incircle $\omega$ so that $DD\equiv BC$ (the third tangency point of the incircle and $\triangle ABC$), i.e. $Q=GH\cap DD$. This motivates constructing the point $X\in \omega$ so that $(XD;GH)=-1$ because then $XX,DD,GH$ concur and thus $QX$ is tangent to $\omega$. Therefore $QXID$ is a kite so $IQ\perp XD$. Thus we want to prove $XD\parallel IM$. Now, the second key observation is recalling that the parallel line to $IM$ through $A$ is nice, in fact, it passes through $D'$, the $D$-antipode on $\omega$ (proof is homotheties at $A$ and $D$ involving extouch point). Thus we wish to show $XD\parallel AD'$. In fact, they are reflections about $I$. It's enough to show that the $X$-antipode $X'$ is collinear with $A$ and $D'$. The proof of this is with phantom points, let $AD'$ hit $\omega$ again at $T$. Then since $A=EE\cap FF$, $D'TEF$ is harmonic. Letting $T'$ be the $T$ antipode, $DT'GH$ is harmonic. However, $DXGH$ is harmonic by definition! So $X=T'$ and we're done.

Let $A'$ be the intersection of the tangents to the incircle at $G$ and $H$, or equivalently the reflection of $A$ over $I$. Let $D$ be the $A$-intouch point, $P$ its antipode with respect to the incircle, and $D'$ be the $A$-extouch point. Because the polar of $A'$ is $\overline{GH}$ by construction, $Q$ lies on the polar of $A'$, hence $A'$ lies on the polar of $Q$. Since $D$ also lies on the polar of $Q$, it follows that $\overline{PD} \perp \overline{IQ}$, so it suffices to show that $\overline{PD} \parallel \overline{IM}$. This is clear by composing a $-1$ ratio homothety at $I$, a homothety at $A$ sending $P$ to $D'$ (equivalently the incircle to the $A$-excircle), and a $\tfrac{1}{2}$ ratio homothety at $D$. $\blacksquare$

yawn bashing is still better We complex bash, setting the unit circle as the incircle. Define $D$ as before; let $d,e,f$ lie on the unit circle, so $b=\frac{2df}{d+f}$ and $c=\frac{2de}{d+e}$. Thus $$m=\frac{e+f}{2}=\frac{d(de+df+2ef)}{(d+e)(d+f)}.$$Further, $g=-e,h=-f$. Since $Q=\overline{GH} \cap \overline{DD}$, by the complex intersection formula $$q=\frac{ef(2d)-d^2(-e-f)}{ef-d^2}=\frac{d(de+df+2ef)}{ef-d^2}.$$It suffices to show that $$\frac{q}{m}=\frac{(d+e)(d+f)}{ef-d^2}$$is pure imaginary, but this is clear since its conjugate is its negative: $$\frac{(\frac{1}{d}+\frac{1}{e})(\frac{1}{d}+\frac{1}{f})}{\frac{1}{ef}-\frac{1}{d^2}}=\frac{(e+d)(f+d)}{d^2-ef}=-\frac{(d+e)(d+f)}{ef-d^2}.~\blacksquare$$

yooo wait Let the reflection of $Q$ across $I$ be $Q'$ and note that $Q'$ lies on $EF$. Also let $D$ be the antipode of the tangency point of the incircle with $BC$, note that $Q'D$ is tangent to the incircle. So $AD$ is the polar of $Q'$ so the polar of $Q$ is parallel to $AD$, now by homothety the polar of $Q$ is parallel to $IM$ and $IQ\perp IM$, done. wait oops post #4 is easier than this i guess . i'm just bad

Let $D$ be the tangent point to the incenter at $\overline{BC}$. Let $Q' \in EF$ be the reflection of $Q$ over $I$ and $D'$ be the point on the incircle such $DD'$ is a diameter. Then, since $Q'D'$ is a tangent, and since $EF$ is the polar of $A$, it follows that $AD'$ is the polar of $Q'$ and thus $AD' \perp IQ'$. It is well-known that $AD' \parallel IM$, which gives the result.

Attachments:

Reflect line $BC$ over $I$ to get line $KL$, where $K$ lies on $AB$ and $L$ lies on $AC$. Clearly, $KL$ is tangent to the incircle $\omega$ of $\triangle ABC$; let $Z$ be this point of tangency, so that $I$ is the midpoint of $DZ$. Finally, let $N=AZ\cap BC$, $X=KL\cap EF$, and define $Y$ to be the reflection of $M$ over $I$. Due to a homothety centered at $A$ mapping $KL$ to $BC$, we find that $N$ is the point of tangency between the $A$-excircle of $\triangle ABC$ and $BC$. Since $D$ is the point of tangency between $\omega$ and $BC$, it follows that $BD=CN$. Chasing lengths along $BC$, it then follows that $MD=BM-BD=CM-CN=MN$, implying that $M$ is the midpoint of $DN$. Combined with the fact that $I$ is the midpoint of $CZ$, it follows that the segment $IM$ is a midline of $\triangle DZN$, so that $IM\parallel ZN$ and thus $ZW\parallel YM$, where $J=IX\cap AZ$. Since $X=IJ\cup YZ$, we then get that $\triangle IXY\sim\triangle JXZ$. Let $W\neq Z$ be the second intersection of $AZ$ and $\omega$, and let $W'\neq Z$ be the unique point on $\omega$ such that $XW'$ is tangent to $\omega$. Then since $AE$, $AF$, $XW'$, and $XZ$ are all tangents to $\omega$, $Z$ lies on $AW$, and $X$ lies on $EF$, it follows that $EZFW$ and $EZFW'$ are both harmonic, forcing us to have $W=W'$. Hence $XZ$ and $XW$ are both tangent to $\omega$, so that $XZ=XW$. Since $ZW\parallel YM$ and $X=ZW\cap YM\cap IJ$, there exists a unique homothety centered at $X$ mapping $ZW$ to $YM$, which also sends $J$ to $I$. It is easy to check that $I$ is the midpoint of $YM$, so homothety implies that $J$ is the midpoint of $ZW$. Since $XZ=XW$, as found above, it follows that $ZW\perp JX$. Applying homothety, we then find that $YM\perp IX$, so that $\angle XIY=90^\circ$. Because $KL$ and $EF$ are the respective reflections of $GH$ and $BC$ over $I$, we must have that $X$ is the reflection of $Q$ about $I$. Similarly, $Y$ is by definition the reflection of $M$ about $I$. Hence $\angle MIQ=\angle XIY=90^\circ$, so we are done. $\blacksquare$ - Jörg

Let $D$ be the other intouch point, and let $\omega$ be the incircle. Reflect $GH$ and $BC$ over $I$ to get $EF$ and $\ell$, which will intersect at $Q'$. Then it suffices to prove that $IQ'\perp IM$. Since $EF$ is the polar of $A$ with respect to $\omega$, $A$ is on the polar of $Q'$ with respect to $\omega$. If $\ell$ is tangent to $\omega$ at $X$, then $X$ is also on the polar of $Q'$ with respect to $\omega$, so we just want to prove that $AX\parallel IM$. Let line $AX$ hit $BC$ at $Y$. Then $Y$ is the tangency point of the $A$-excircle. Then $DM=MY$ and $DI=IX$, so line $IM$ is the $D$-midline of triangle $DXY$, as desired.

Let $K$ be the $A-$extouch point and let $D$ be the point where the incircle touches $BC$. We also consider $D'$ i.e. the antipode of $D$ w.r.t $I$. By the 'Diameter of incircle' lemma we know $BD=KC$ but $BM=MC$ too which makes it clear that $M$ is also the midpoint of $DK$. And we also know $A-D'-K$ which implies $MI \parallel KD'$ thus $MI \parallel AD'$. Also let $A'$ be the reflection of $A$ over $I$ which is basically the intersection of tangents from $H$ and $G$. Thus $HG$ is the polar of $A'$ and $Q \in a'$. By La Hire's theorem we have $A' \in q$ but as $QD$ is tangent to the incircle therefore $D \in q$. Hence $A'D$ is the polar of $Q$ which proves $A'D \perp QI$. As $AI=IA'$ and $DI=ID'$ we have $AD'A'D$ a parallelogram. Thus $A'D \parallel AD'$ which implies $QI \perp AD'$. But as $MI$ was parallel to $AD'$ we also have $QI \perp MI$. $\blacksquare$

Nice Problem Let $X$ intersection of tangent from $G$ and $H$ to incircle and defines $D$ as touch point for $BC$ Note from $XG \parallel AE$ and $XH \parallel AF$ we get $X$ is reflection of $A$ about $I$ Polar of $Q$ passes through $D$ and $X$ also (As line $GH$ is polar of $X$).hence $XD$ is polar of $Q$ and $XD \perp IQ$ Let $O$ be intersection of $ID$ with incircle then as $ID=IO$ we get $XD \parallel AO$ its well known $AO \parallel IM$ which give us $QI \perp IM$ $\blacksquare$

Let the altitude of $I$ onto $BC$ be $D$, and the antipode of $D$ wrt the incircle be $D'$. Then let the reflection of $Q$ over $I$ be $Q'$. Notice that the pole of $EF$ is $A$ and the pole of $\overline{D'D'}$ is $Q$. By reflection, we have $\overline{EF} \cap \overline{D'D'} = Q'$ and by La Hire's, $Q'$ is the pole of $AD'$. This means that the pole of $Q$ is parallel to $AD'$, and $AD' \parallel IM$ finishes.

Here's a fairly quick complex bash. Let the incircle be the unit circle, so $g=-e$ and $h=-f$, and in particular \begin{align*} q &= \frac{2def+d^2e+d^2f}{ef-d^2} \\ m &= \frac{df}{d+f} + \frac{de}{d+e}. \end{align*}Hence $$\frac qm = \frac{(d+e)(d+f)}{ef-d^2}$$is the negative of its conjugate and thus imaginary, as needed.

Let $D$ be the tangency of the incircle with $\overline{BC}$ and denote $D'$ as the antipode of $D$ with respect to the incircle. Then, reflect $Q$ over $I$ to point $Q'$, noting that $Q'$ is obviously tangent to the incircle and it lies on $\overline{EF}$. Since $A$ is the pole of $\overline{EF}$, we must have $\overline{AD'}$ be the polar of $Q'$. The well-known property that $\overline{AD'} \parallel \overline{IM}$ finishes. $\square$

let $T$ be antipode of $D$, let $P=TT\cap EF$ clearly $I$ is the midpoint of $PQ$. now $AT\parallel IM$ by homothety at $D$ and $AT$ is the polar of $P$ implies the conclusion