Did anyone manage to bash this? I tried with vectors for a while, but it was too messy

I length bashed (a), but didn't have time for (b).

mathwizard888 wrote: I length bashed (a), but didn't have time for (b). Same.

It was easy for (A) but I couldn't find X in vectors. (A) is tricky with all the factoring though.

Bary bashed both parts gggg

And who is surprised.

Diagram for the problem. Solved through massive bashing, need elegant solutions please!

Guys, if you're taking Olympiad Geometry, this was the polar to the incircle/excircle configuration that dragon96 posted maybe Week 5. Synthetic sketch: (a) BUVC is cyclic since they are equidistant to M, so then $\angle VCB=1/2 \angle VMB=1/2 \angle C$ (b) you can show that IX is the I-symmedian of triangle IBC since the midpoint of arc BC not containing A is its circumcenter

@ABCDE Can you elaborate how MV=MU implies BVUC cyclic? I think that would involve MV= BC/2 which would solve part a anyways.

Yes, MV=MU=MB=MC. You can do this with similar triangles and some s-c's and stuff. Thank you so much dragon96 for starting Configuration Practices.

on ABCDE's method.

here's part b in it's full synthetic glory

and robin I thought you stopped bashing

If you look at my post at 5:42, you will see that I posted a few more details than fclvbfm934 at 6:43. Also, for although I didn't write anything for (b), I did think of symmedians when I was going through the advanced methods I've heard of but never learned. I'm wondering if anyone found a solution using the $\v{S}$-point, because at AwesomeMath, we learned about the $\v{S}$-point, but I forgot everything about it except the definition, from which I got that $X$ is the reflection of $\v{S}_a$ over $O$. Maybe this means we can apply a transformation to $AI$ to define it differently. Darjn symmedians are too advanced for me. If no one posts an elementary synthetic solution for (b), I'm going to try it again (well I'm going to try it again anyways cuz its fun).

Hmm does anyone know if perspectivity was a possible solution?

"we will bary you" - Schindler - Khruschev

The full configuration for part (a) can be stated as follows: Let $M$ be the midpoint of $\overline{BC}$ of triangle $ABC$ with intouch triangle $DEF$ and incenter $I$. Let ray $BI$ meet line $EF$ at $K$. Then $\angle BKC = 90^{\circ}$ and $\overline{MK} \parallel \overline{AB}$. In fact, you don't need intouch lengths to prove this, as this lemma can be derived entirely from angle chasing with a well-chosen cyclic quadrilateral. (Try it!) This happens to be one of the practice problems in the first chapter of my textbook draft. I personally call this the "Iran incenter lemma" because I learned it from failing to solve a certain Iran TST 2009 #9 back in 2012 (one of my favorite tests of configuration recognition). It's also showed up in a lot of other places (eg a WOOT PO #3 from some time ago). I can't say it falls under "well-known" but it certainly comes close. By the way, I may or may not have been responsible for dragon96 sharing this configuration with you Olympiad geometry students I'll have to check with him later to see of I in fact generated this coincidence...

For the angle chase, you can easy angle chase $DIFK$ as a cyclic quadrilateral. This means that $CDIFK$ is cyclic. Then $\angle IKC=90\text degrees$, so by some right triangle stuff, we're done. And thanks if you started configurations.

Solution for part b) Let R be the intersection of BI and CI,and Y be the middle of the arc BAC,and N be the orthocenter of RVU,so we have FIEN is a parallelogram,so we have to prove that I,N and R are collinear,so if we prove that triangle R'NI is similar to YIX,we are finished.Now,we have that <NRI=<IYX,by trivial angle chase,and RI/XY=RN/ YI= 2*(sin A/2)^2,easily from sine law,but this can be easily be solved withuot this little trig,or totaly elementary,so the similarity is proven and we are finished

(a) This is just the statement of the Iran Lemma. (b) Let $Q$ be the midpoint of $\overline{UV}$. Since $\angle MUV=\angle AFE=\angle AEF=\angle MVU$, $Q$ is the foot of the altitude from $M$ to $\overline{EF}$. Let $R=\overline{AI} \cap \overline{EF}$, $S=\overline{MQ} \cap \overline{AX}$, and $Y=\overline{AI} \cap (ABC)$. Notice that $\overline{AX} \parallel \overline{EF}$ and $\overline{AI} \parallel \overline{MQ}$. We have $\angle EAF=\angle BXC$, $AE=AF$, and $XB=XC$, so $\triangle AFE \sim \triangle XBC$. Since $I$ and $Y$ are corresponding arc midpoints in $(AFE)$ and $(XBC)$, we have $AFIE \sim XBYC$. We have $\angle XAI=\angle XSQ=90^\circ$ and \[\frac{XA}{XS}=\frac{XY}{XM}=\frac{AI}{AR}=\frac{\operatorname{dist}(I,\overline{AX})}{\operatorname{dist}(Q,\overline{AX})},\]so $\triangle XAI \sim \triangle XSQ$, which means $I$, $Q$, and $X$ are collinear.

(a) Is just Iran Lemma. For (b), notice how $\angle BVC=\angle BUC=90.$ Thus we get that $MB=MV=MU=MC,$ which implies that the midpoint of $UV$ is the projection of $M$ on $EF$. Now the problem follows from Peru TST D1P2 (see my post): https://artofproblemsolving.com/community/u930637h2857859p27167407 .

easy and cute barybash (also my new favorite geo problem )

U and V are just the intersections of EF with the interior B and C angle bisectors. Then homothety to theidpoint of the arc BCA, and the B C excenters

I think I will post this one. Times have changed man... EGMO revolutionised geometry I guess... Solution: Denote $X$ as $X_1$ and $X_2$ as the second arc midpoint of $BC$. Let $X_1I$ hit $\odot(ABC)$ and $\odot(BIC)$ at $Q$ and $Q'$ respectively. [asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.525961355717461, xmax = 11.0, ymin = -6.45, ymax = 5.8; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen dcrutc = rgb(0.8627450980392157,0.0784313725490196,0.23529411764705882); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); draw((-0.6797360363214043,2.685040650406503)--(-0.9,-2.38)--(8.38,-2.34)--cycle, linewidth(0.6) + qqwuqq); /* draw figures */ draw((-0.6797360363214043,2.685040650406503)--(-0.9,-2.38), linewidth(0.6) + qqwuqq); draw((-0.9,-2.38)--(8.38,-2.34), linewidth(0.6) + qqwuqq); draw((8.38,-2.34)--(-0.6797360363214043,2.685040650406503), linewidth(0.6) + qqwuqq); draw(circle((3.73001739833879,-0.0440364145992922), 5.1859219991471175), linewidth(0.6) + blue); draw(circle((1.0867347921461603,-0.4695609738113125), 1.9018578468581937), linewidth(0.6) + dcrutc); draw((-0.8133272704369741,-0.38693272205436136)--(3.9415915207779273,2.275661823499158), linewidth(0.6) + yqqqyq); draw((3.74,-2.36)--(3.9415915207779273,2.275661823499158), linewidth(0.6) + yqqqyq); draw((3.74,-2.36)--(-0.7898680181607022,0.15252032520325165), linewidth(0.6) + yqqqyq); draw((3.707664493920653,5.141837410408611)--(3.752370302756928,-5.229910239607197), linewidth(0.6) + yqqqyq); draw((-0.2611228965335519,-3.3553180589002345)--(3.707664493920653,5.141837410408611), linewidth(0.6) + yqqqyq); draw((-0.6797360363214043,2.685040650406503)--(1.0867347921461603,-0.46956097381131284), linewidth(0.6) + yqqqyq); draw(circle((3.7523703027569284,-5.229910239607198), 5.455871865045269), linewidth(0.6) + yqqqyq); draw((-0.9,-2.38)--(3.9415915207779273,2.275661823499158), linewidth(0.6) + yqqqyq); draw((-1.6089805852132641,-6.241075143989156)--(-0.2611228965335519,-3.3553180589002345), linewidth(0.6) + yqqqyq); /* dots and labels */ dot((-0.6797360363214043,2.685040650406503),dotstyle); label("$A$", (-0.9720473022912114,3.0713821138211372), NE * labelscalefactor); dot((-0.9,-2.38),dotstyle); label("$B$", (-1.444242424242431,-2.5499883855981396), NE * labelscalefactor); dot((8.38,-2.34),dotstyle); label("$C$", (8.651738992714597,-2.437560975609754), NE * labelscalefactor); dot((1.0867347921461603,-0.46956097381131284),linewidth(4.pt) + dotstyle); label("$I$", (1.3889283074648866,-0.6612078977932625), NE * labelscalefactor); dot((3.74,-2.36),linewidth(4.pt) + dotstyle); label("$M$", (3.9747587371977557,-2.9322415795586503), NE * labelscalefactor); dot((3.850131981839298,0.17252032520325167),linewidth(4.pt) + dotstyle); label("$N$", (4.0647006651884645,0.5754936120789783), NE * labelscalefactor); dot((-0.7898680181607022,0.15252032520325165),linewidth(4.pt) + dotstyle); label("$P$", (-1.264358568261014,0.4180952380952386), NE * labelscalefactor); dot((1.0949324305101018,-2.3714011533167665),linewidth(4.pt) + dotstyle); label("$D$", (0.9392186675113441,-2.9322415795586503), NE * labelscalefactor); dot((2.009215664284388,1.1935963579511324),linewidth(4.pt) + dotstyle); label("$E$", (1.6812395734346892,1.6772822299651564), NE * labelscalefactor); dot((-0.8133272704369741,-0.38693272205436136),linewidth(4.pt) + dotstyle); label("$F$", (-1.3093295322563683,-0.6612078977932625), NE * labelscalefactor); dot((3.9415915207779273,2.275661823499158),linewidth(4.pt) + dotstyle); label("$U$", (4.042215183190787,2.464274099883855), NE * labelscalefactor); dot((-0.3176753872531194,-0.10938434350014678),linewidth(4.pt) + dotstyle); label("$V$", (-0.2749973603632206,0.21572590011614465), NE * labelscalefactor); dot((3.707664493920653,5.141837410408611),linewidth(4.pt) + dotstyle); label("$X_1$", (3.7948748812163386,5.319930313588848), NE * labelscalefactor); dot((3.752370302756928,-5.229910239607197),linewidth(4.pt) + dotstyle); label("$X_2$", (3.7274184352233073,-5.765412311265965), NE * labelscalefactor); dot((-0.2611228965335519,-3.3553180589002345),linewidth(4.pt) + dotstyle); label("$Q$", (-0.7696779643121173,-3.764204413472703), NE * labelscalefactor); dot((-1.6089805852132641,-6.241075143989156),linewidth(4.pt) + dotstyle); label("$Q'$", (-2.2762052581564847,-6.305063879210216), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] The solution won't prove the first part since it is a well-known lemma in EGMO. The proof is just angle chasing with midpoint theorem(people are citing Iran lemma too, so that's that...). For the second part, by the incenter-excenter lemma $Q'IBC$ is harmonic quadrilateral. Note that tangent at $I$(call it $\ell$) to $\odot(BIC)$ is parallel to $EF$. Let $EF \cap \ell = \infty$. Finally \[-1 = (Q',I;B,C) \overset{I}{=} (UV \cap X_1I, \infty; U,V) \]which proves the bisection as desired. $\blacksquare$

Hopefully this works. The first part is trivial by Iran Lemma. To prove the second part, let the midpoint of arc $BC$ not containing $A$ be $M_A$, and similarly define $M_B$ and $M_C$. Since $XBM_AC$ is a kite, then \[ (X, M_A; B, C) = -1 \]Projecting through $I$ onto the circle $(ABC)$ again gives \[ (K, A; M_B, M_C) = -1 \]where $K$ is the intersection of $IX$ with $(ABC)$ where $K \neq X$. This implies that $KA$ is the $K$-symmedian of $\triangle KM_BM_C$. Now, since $\angle M_CKA = \angle M_CCA = \frac{1}{2} \angle BCA$ and \[ \angle XKM_B = \angle XKC - \angle M_BKC = \angle XKM_A - \angle CKM_A - \frac{1}{2} \angle ABC = \frac{1}{2} \angle BCA \]Then $KX$ is the median so $IX$ bisects $M_BM_C$. Since $EF \parallel M_BM_C$ (consider a homothety from incircle to circumcircle), then $I$ must be the center of the homothety sending $M_BM_C$ to $UV$, so $IX$ also bisects $UV$, as desired. $\blacksquare$

(a) Phantom $V' = CI\cap EF$. Then $$\measuredangle V'IB = \measuredangle CIB = \measuredangle EFB = \measuredangle V'FB$$ implying $V'FIB$ is cyclic, so $V'$ is the foot from $B$ to $CI$, which lies on $MP$ since the reflection of $B$ over $CI$ lies on $CA$. (b) Extending $CI$ and $BI$ to hit $(ABC)$ at the arc midpoints of $\widehat{AB}$ and $\widehat{AC}$, we see by converse of Reim that $BCUV$ is cyclic, so since $IX$ is a symmedian in $\triangle BIC$ it is a median in $\triangle UIV$.

Part (a) is a direct application of the Iran lemma. Let $T$ be the intersection of $IX$ and $EF$, and let $S$ be the intersection of $BC$ and $IX.$ We wish to show that $$(VU;T\infty)=-1.$$Suppose that the line parallel to $EF$ (and thus perpendicular to $AI$) at $I$ intersects $BC$ at $R$. Then, we have $$(VU;T\infty)=^I(CB;SR).$$ Let $IX$ intersect $(ABC)$ again at $K$, and let the arc midpoint of $BC$ be $L$ (so $L$ is the antipode of $X$). Then, obviously $\angle XKL=90$. Now, note that $(IKL)$ is tangent to $IR$ since $IR$ is perpendicular to $AL$, $(IBC)$ is tangent to $IR$ as well due to Fact 5. Hence, the radical center of $(IBC),(IKL),(BCKL)$ is $R$, which means that $R,K,L$ are collinear. Hence, projecting from $K$ onto $(ABC)$, we have $$(CB;SR)=(CB;XL)$$which is clearly a harmonic quadrilateral, hence done.

Nice. Part (a) is just Iran Lemma. For part (b), let the $A$-mixtillinear incircle touch $AC,AB, \Gamma$ at $Q,R,T,$ respectively, and let $Y = QR \cap BC.$ Also set $W = \overline{XIT} \cap \overline{BC}$ and $S = \overline{XIT} \cap \overline{UV}.$ Projecting through $I,$ we see that$$(U,V;S;P_{\infty}) = (B,C; W,Y).$$If we project this cross-ratio from $T$ onto $\Gamma,$ we get$$(B,C; W,Y) = (B,C;X, YT \cap \Gamma).$$It thus suffices to show that $\angle YTI = 90^\circ.$ Now let $\omega$ be the circle with diameter $AI,$ and let $Z$ be the second intersection of $\omega$ and $\Gamma.$ By Radical Center on $\Gamma, \omega,$ and $(BIC),$ we have that $A,Z,Y$ are collinear. Thus $\angle YZI = 90^\circ.$ Therefore, we just need to show that $YZIT$ is cyclic. But we have$$\measuredangle YZT = \measuredangle AZT = \measuredangle ABT = \measuredangle RBT = \measuredangle RIT = \measuredangle YIT$$since $BRIT$ is cyclic (well-known), so we are done! Ok yeah this solution is overkill.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.082, xmax = 17.434, ymin = -3.3, ymax = 10.076; /* image dimensions */ pen qqffff = rgb(0,1,1); pen ffqqff = rgb(1,0,1); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((0.657455195014884,8.02160090233444)--(0,0)--(10.269245168233319,0)--cycle, linewidth(0.4) + qqffff); draw((1.1924725648728778,3.2899547576064165)--(2.899227532861332,2.671326653502196)--(5.554052902138825,5.117462973939788)--cycle, linewidth(0.8) + ffqqff); draw((2.899227532861332,2.671326653502196)--(0,0)--(10.269245168233319,0)--cycle, linewidth(0.8) + ffqqff); draw((1.1129566831763562,3.070575577741907)--(1.3323358630408657,2.9910596960453857)--(1.4118517447373873,3.210438875909895)--(1.1924725648728778,3.2899547576064165)--cycle, linewidth(0.4) + qqwuqq); draw((5.382446025507123,4.959345676047648)--(5.5405633233992635,4.787738799415946)--(5.7121702000309655,4.945856097308086)--(5.554052902138825,5.117462973939788)--cycle, linewidth(0.4) + qqwuqq); draw(arc((2.899227532861332,2.671326653502196),0.66,-50.07704238699136,-19.92348669144255)--(2.899227532861332,2.671326653502196)--cycle, linewidth(0.8) + blue); draw(arc((2.899227532861332,2.671326653502196),0.66,-137.34276163324904,-107.18920593770022)--(2.899227532861332,2.671326653502196)--cycle, linewidth(0.8) + blue); draw(arc((2.899227532861332,2.671326653502196),0.66,42.65723836675097,72.81079406229978)--(2.899227532861332,2.671326653502196)--cycle, linewidth(0.8) + blue); draw(arc((2.899227532861332,2.671326653502196),0.66,129.92295761300863,160.07651330855748)--(2.899227532861332,2.671326653502196)--cycle, linewidth(0.8) + blue); draw(arc((10.269245168233319,0),0.66,160.07651330855745,180)--(10.269245168233319,0)--cycle, linewidth(0.8) + qqwuqq); draw(arc((5.554052902138825,5.117462973939788),0.66,-157.2662483246916,-137.34276163324904)--(5.554052902138825,5.117462973939788)--cycle, linewidth(0.8) + qqwuqq); draw(arc((0,0),0.66,0,42.657238366750974)--(0,0)--cycle, linewidth(0.8) + red); draw(arc((1.1924725648728778,3.2899547576064165),0.66,-19.92348669144254,22.733751675308415)--(1.1924725648728778,3.2899547576064165)--cycle, linewidth(0.8) + red); /* draw figures */ draw((0.657455195014884,8.02160090233444)--(0,0), linewidth(0.4) + qqffff); draw((0,0)--(10.269245168233319,0), linewidth(0.4) + qqffff); draw((10.269245168233319,0)--(0.657455195014884,8.02160090233444), linewidth(0.4) + qqffff); draw(circle((5.134622584116659,3.616906431243753), 6.280633822608452), linewidth(0.4)); draw(circle((2.899227532861332,2.671326653502196), 2.671326653502196), linewidth(0.4)); draw((0.23682829773862535,2.8895384829919526)--(5.554052902138825,5.117462973939788), linewidth(0.4)); draw((1.1924725648728778,3.2899547576064165)--(2.899227532861332,2.671326653502196), linewidth(0.8) + ffqqff); draw((2.899227532861332,2.671326653502196)--(5.554052902138825,5.117462973939788), linewidth(0.8) + ffqqff); draw((5.554052902138825,5.117462973939788)--(1.1924725648728778,3.2899547576064165), linewidth(0.8) + ffqqff); draw((2.899227532861332,2.671326653502196)--(0,0), linewidth(0.8) + ffqqff); draw((0,0)--(10.269245168233319,0), linewidth(0.8) + ffqqff); draw((10.269245168233319,0)--(2.899227532861332,2.671326653502196), linewidth(0.8) + ffqqff); draw((0,0)--(1.1924725648728778,3.2899547576064165), linewidth(0.4)); draw((10.269245168233319,0)--(5.554052902138825,5.117462973939788), linewidth(0.4)); draw((2.072865248021589,0)--(5.134622584116659,9.897540253852206), linewidth(0.4)); draw((2.0728652480215892,3.658840514061978)--(5.134622584116659,0), linewidth(0.4)); /* dots and labels */ dot((0.657455195014884,8.02160090233444),linewidth(1pt) + dotstyle); label("$A$", (0.736,8.074), NE * labelscalefactor); dot((0,0),linewidth(1pt) + dotstyle); label("$B$", (-0.232,-0.308), NE * labelscalefactor); dot((10.269245168233319,0),linewidth(1pt) + dotstyle); label("$C$", (10.35,0.044), NE * labelscalefactor); dot((2.899227532861332,2.671326653502196),linewidth(4pt) + dotstyle); label("$I$", (2.98,2.838), NE * labelscalefactor); dot((2.899227532861332,0),linewidth(1pt) + dotstyle); label("$D$", (2.98,0.044), NE * labelscalefactor); dot((5.134622584116659,9.897540253852206),linewidth(1pt) + dotstyle); label("$X$", (5.224,9.548), NE * labelscalefactor); dot((0.23682829773862535,2.8895384829919526),linewidth(1pt) + dotstyle); label("$E$", (-0.012,2.926), NE * labelscalefactor); dot((4.61085164716268,4.72226033869374),linewidth(1pt) + dotstyle); label("$F$", (4.586,4.906), NE * labelscalefactor); dot((1.1924725648728778,3.2899547576064165),linewidth(1pt) + dotstyle); label("$V$", (1.11,3.542), NE * labelscalefactor); dot((5.554052902138825,5.117462973939788),linewidth(1pt) + dotstyle); label("$U$", (5.642,5.17), NE * labelscalefactor); dot((5.134622584116659,0),linewidth(1pt) + dotstyle); label("$M$", (5.224,0.044), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Part $(a)$ it´s Iran lemma For $(b)$ note that $IX$ is the symmedian of $\triangle IBC$ because if $N$ is the center of $IBC$ then $\angle NBX=\angle NCX$; so focus in cyclic quadrilateral $VUCB$, $\triangle IBC\sim \triangle IVU$ and as $\angle CIM=\angle UIX$, then $IX$ bisects $UV$, as desired

The bulk of our solution uses relevant facts from the Iran Lemma, which trivializes part (a). An alternative formulation to part (b) is to denote $K$ as the midpoint of $UV$, and show that $IK$ and $IX$ represent the same line. Note that $\triangle BIC \sim \triangle IVU$ with opposite orientation. Incenter-Excenter Lemma tells us $XI$ is a symmedian in $\triangle BIC$, so it must be a median in $\triangle IUV$, as desired.

Does this work? (a) To prove that $I$ lies on ray $CV$, we will use the properties of the incenter. Since $E$ and $F$ are the tangency points of the incircle $\gamma$ with sides $AC$ and $AB$ respectively, $ME = MF = s - b$, where $s$ is the semi-perimeter of triangle $ABC$. This is because the tangents from a point outside the circle to the circle are equal in length. Now, let's denote $D$ as the tangency point of $\gamma$ with $BC$. By the Midpoint Theorem, $MD$ is parallel to $AB$, and $MD = \frac{1}{2} (s - b)$. Now, consider triangles $IME$ and $IFD$. They share the angle $\angle I$, and both triangles have a right angle at $I$. Also, $IM = IF$ (inradius property), and $ME = FD$ as established earlier. Therefore, by SAS congruence, triangles $IME$ and $IFD$ are congruent. So, $\angle MIE = \angle FID$. Now, since $X$ is the midpoint of arc $BC$ not containing $A$, we have $\angle XIB = 90^\circ$. Therefore, $\angle MIC = 90^\circ - \angle MIE = 90^\circ - \angle FID = \angle FIB = \angle FCB$. So, $MICF$ is cyclic. Now, we have $\angle CVF = \angle CMF = \angle CMI = \frac{1}{2} \angle CIM$. But we know $\angle CIM = \frac{1}{2} \angle CIB = \frac{1}{2} \angle CAB$. Hence, $\angle CVF = \frac{1}{2} \angle CAB = \angle CIX$. Therefore, $CV$ is parallel to $IX$. However, since $X$ is between $C$ and $I$ on the arc, the ray $CV$ must also pass through $I$. Thus, $I$ lies on ray $CV$. (b) To prove that line $XI$ bisects $\overline{UV}$, we will use properties of similar triangles. Since $XI$ is the angle bisector of $\angle BXC$, it divides $\overline{BC}$ proportionally. That is, $\frac{BI}{CI} = \frac{BX}{CX}$. Now, let's denote $H$ as the point of intersection of $XI$ and $UV$. We need to prove that $HU = HV$. From the similar triangles $BXI$ and $CXH$, we have $\frac{BX}{CX} = \frac{BI}{CI} = \frac{IU}{IH}$ (because $BI$ bisects $\angle UIB$ and $CI$ bisects $\angle HIC$). Similarly, from the similar triangles $BXI$ and $CXH$, we have $\frac{BX}{CX} = \frac{BI}{CI} = \frac{IV}{IH}$ (because $BI$ bisects $\angle IVB$ and $CI$ bisects $\angle HVC$). Thus, $IU = IV$. Therefore, line $XI$ bisects $\overline{UV}$. This completes the proof.

Massive overkill for part (b): Note that $T = \overline{XI} \cap \Gamma$ is the mixtilinear touchpoint. In particular, let the mixtilinear incircle touch $\overline{AB}$ and $\overline{AC}$ at $T_B$ and $T_C$, and let $M_B = \overline{TT_B} \cap \Gamma$ and similarly $M_C$ be arc midpoints. Then, $\overline{T_BT_C} \parallel \overline{UV} \parallel \overline{M_BM_C}$. Then $\overline{IT}$ bisects $\overline{M_BM_C}$, but taking a homothety at $I$ yields $\overline{IT}$ bisects $\overline{UV}$, as needed.

Part a is just Iran lemma. For part b, Since $MV=MU$, note that the midpoint of $UV$ is the foot of the perpendicular from $M$ to $EF$. We employ barycentric coordinates with reference triangle $ABC$. Then note that this perpendicular must be parallel to $AI$, which has equation $cy-bz=0$. Then the parallel through $M$ will have equation \[(b-c)x+(b+c)y-(b+c)z=0\]since these two lines won't intersect. Now note that $E=(s-c:0:s-a)$ and $F=(s-b:s-a:0)$. Then line $EF$ has equation \[(s-a)x-(s-b)y-(s-c)z=0.\]Let's intersect these two lines. Then \[(b-c)(s-a)x+(b+c)(s-a)y-(b+c)(s-a)z=0\]\[-(b-c)(s-a)x+(b-c)(s-b)y+(b-c)(s-c)z=0.\]Adding these and multiplying by $2$, \[\Big((b+c)(b+c-a)+(b-c)(a+c-b)\Big)y+ \Big(-(b+c)(b+c-a)+(b-c)(a+b-c)\Big)z=0\]\[\Big(4bc-2ac\Big)y+ \Big(2ab-4bc\Big)z=0.\]So we let $Y$ be the foot of the perpendicular from $M$ to $EF$, and set \[Y=\left(t:(a-2c)b:(a-2b)c\right).\]Then note that \[(b-c)t+(b+c)((a-2c)b-(a-2b)c)=0\]\[(b-c)t+(b+c)(ab-ac)=0\]\[t=-a(b+c).\]So \[Y=\left(-a(b+c):(a-2c)b:(a-2b)c\right).\] It is well known that $XI$ passes through the $A$-mixtilinear touch point, say $T$. We know that $I=(a:b:c)$. Note that $T$ is the $\sqrt{bc}$ inverse of the $A$-extouch point, which is $(0:s-b:s-c)$, so $T=(a^2t:b^2(s-c):c^2(s-b))$. Plugging this into the circumcircle equation, \[a^2b^2c^2(s-c)(s-b)+a^2b^2c^2(s-b)t+a^2b^2c^2(s-c)t=0,\]so $(s-c)(s-b)+at=0$, so $t=-\frac1a (s-c)(s-b)$. Thus we have \[T=(-a(s-c)(s-b):b^2(s-c):c^2(s-b)).\] Thus it suffices to verify the following. The equations below will be iff each other, so I will omit the iff symbol. \[0=\begin{vmatrix} a & b & c \\ -a(b+c) & (a-2c)b & (a-2b)c \\ -a(s-b)(s-c) & b^2(s-c) & c^2(s-b) \end{vmatrix}\]\[0=\begin{vmatrix} a & b & c \\ -a(b+c) & (a-2c)b & (a-2b)c \\ -a(a+b-c)(a-b+c) & 2b^2(a+b-c) & 2c^2(a-b+c) \end{vmatrix}\]\[0=\begin{vmatrix} a & b & c \\ a(a+b+c) & 2bc & 2bc \\ -a(a+b-c)(a-b+c) & 2b^2(a+b-c) & 2c^2(a-b+c) \end{vmatrix}\]\[0=\begin{vmatrix} 1 & 1 & 1 \\ a+b+c & 2c & 2b \\ -(a+b-c)(a-b+c) & 2b(a+b-c) & 2c(a-b+c) \end{vmatrix}\]\[0=\begin{vmatrix} 1 & 0 & 0 \\ a+b+c & -a-b+c & -a+b-c \\ -(a+b-c)(a-b+c) & (a+b+c)(a+b-c) & (a+b+c)(a-b+c) \end{vmatrix}\]\[0=\begin{vmatrix} -a-b+c & -a+b-c \\ (a+b+c)(a+b-c) & (a+b+c)(a-b+c) \end{vmatrix}\]which is true. $\blacksquare$

sketch: part a is just a well known lemma (easy angle chasing) For part b it can be easly done by the nine point circle