Amazing Problem! Let the $A-$excircle touch $AB,AC$ at $A_1A_2$, and now invert through a circle centered at $A$ with radius $AA_1$ and then the complete problem turns into a simple trig bash!

I baried it

Trivial after noting that $M$ is the center of $(AJKL) $ where K, L are touch points of the excircle with the sides

Stewart's theorem FTW

We first prove a lemma. Lemma. Let $ABC$ be a triangle with circumcircle $\Omega$. Denote by $I$ the incenter of $\triangle ABC$, and let $AI$ intersect $\Omega$ for the second time at $M$. Then \[ \frac{MI}{AI} = \frac{a}{2(s-a)}. \] Proof. Let $\ell := AI$ and $r := BM = CM = IM$. Then $AM = r + \ell$, so Ptolemy on cyclic quadrilateral $ABMC$ yields $a(\ell + r) = cr + br$. The desired ratio is then $\tfrac{r}{\ell} = \tfrac{a}{2(s-a)}$ as desired. $\square$ Returning to the original proof, let $N$ be the midpoint of arc $\widehat{BC}$. The crucial claim is that $\triangle CMT\sim\triangle CNB$. Indeed, observe that $\angle CNM = \angle CBA$ while \[ \frac{CN}{NM} = \frac{NI}{NM} = \frac{2NI}{AI} \stackrel{(*)}= \frac{2\cdot BC}{2\cdot BT} = \frac{BC}{BT}. \]Note that in $(*)$ we use the lemma as well as the fact that $BT = s-a$. Hence $\triangle CNM\sim\triangle CBT$, and the result follows by spiral similarity.

AlastorMoody wrote: The side $AB$ of $\Delta ABC$ touches the corresponding excircle at point $T$. Let $J$ be the center of the excircle inscribed into $\angle A$, and $M$ be the midpoint of $AJ$. Prove that $MT = MC$. The incircle of $\triangle ABC $ is tangent to $BC, AB $ at $ D, E $, respectively. The $A-$excircle of $\triangle ABC $ is tangent to $BC, CA, AB$ at $ X, Y, Z $, respectively. $MY=MA$, $CY=CX=BD=BE=TA $ and $\angle MYC=\angle MAC=\angle MAT $. Therefore, $ \triangle MYC=\triangle MAT $. Then $MC=MT $.

djmathman wrote: Indeed, observe that $\angle CNM = \angle CBA$ while \[ \frac{CN}{NM} = \frac{NI}{NM} = \frac{2NI}{AI} \stackrel{(*)}= \frac{2\cdot BC}{2\cdot BT} = \frac{BC}{BT}. \] Please how is NI/NM = 2NI/AI

That's equivalent to $AI = 2NM$. This follows from $M$ being the midpoint of $\overline{AJ}$ and $N$ being the midpoint of $\overline{IJ}$, which means that, for example, \[ AI = AJ - JI = 2(MJ - NJ) = 2MN. \]

Thanks Sorry I have another please: How is BT = s-a and how does spiral similarity imply that MT = MC djmathman wrote: The formula $BT = s-a$ is a relatively well-known result about excircles of a triangle; I encourage you to try to derive it yourself. (If you get stuck, a derivation can be found here.) As for your second claim, one can show that $\triangle CNM\sim\triangle CBT$ implies $\triangle CNB\sim\triangle CMT$ (again, I encourage you to check this yourself); then, since $\triangle CNB$ is isosceles, $\triangle CMT$ is too. Thank you sooo much!

The formula $BT = s-a$ is a relatively well-known result about excircles of a triangle; I encourage you to try to derive it yourself. (If you get stuck, a derivation can be found here.) As for your second claim, one can show that $\triangle CNM\sim\triangle CBT$ implies $\triangle CNB\sim\triangle CMT$ (again, I encourage you to check this yourself); then, since $\triangle CNB$ is isosceles, $\triangle CMT$ is too.

if The verdict is in place then should be triangle ATM=AMC => AT= AC =>Contradiction PROBLEM IS ABSOULUTLY FALSE! HOW THEM SOLVE IT!

arzhang2001 wrote: if The verdict is in place then should be triangle ATM=AMC => AT= AC =>Contradiction PROBLEM IS ABSOULUTLY FALSE! HOW THEM SOLVE IT! SSA congruency isnt a thing

Cute! Lemma: If $AB,AC$ are fixed rays and $D,E \in AB,AC$ respectively and $AD+AE$ is fixed, then the circumcircle of $(ADE)$ passes through a fixed point, moreover that fixed point is midpoint of arc $DE$ Now let $J,K$ be points of tangencies of excircle with $AB,AC$ and let $M,N$ be the midpoints of $AJ,AK$ then clearly $AJKM$ is cyclic which in turn implies $ATMC$ must be cyclic by the above lemma, so $MT=MC$.

Use cosine rule in $\triangle ATM$ and $\triangle AMC$. Let $a,b,c$ be the sidelengths of the triangle and $s$ be the semiperimeter. Let $\theta=\frac{\angle A}{2}$ \[MT^2=(s-b)^2+AM^2-2AM(s-b)\cos\theta\]And \[MC^2=b^2+AM^2-2AM\cdot b \cos\theta\]Hence we want \[(s-b)^2-b^2=2AM(s-b-b)\cos\theta\]upon equating, cancelling and rearranging. However, $2AM=AJ$, and considering the foot of $J$ onto $AC$, $2AM\cos\theta=s$ from a right triangle. Now we expand the LHS, which is \[s^2-2bs+b^2-b^2=s^2-2bs=s(s-2b)\]While the RHS is also $s(s-2b)$ clearly. We are done.