do you have to chat as soon has i get aops up?

DiamondPuppy22 wrote: do you have to chat as soon has i get aops up? I do not understand what you just said, probably you meant to answer somewhere else here we do not chat

Very known result.

Solution 1 (Spiral Similarity): Let the incircle touches $AC,AB$ at $E,F$. Then just notice the spiral similarity $\triangle KBF\stackrel{+}{\sim}\triangle KCE$ thus $$\frac{KB}{KC}=\frac{BF}{CE}=\frac{BD}{DC}$$or $KD$ bisects $\angle BKC$. This immediately implies $K,D,M$ are colinear. Solution 2 (Inversion): Again, let $E,F$ be the other two intouch points. Perform inversion around the incircle. We deduce the following facts. $\triangle A'B'C'$ is medial triangle of $\triangle D'E'F'$. $I$ is orthocenter of $\triangle A'B'C'$. $M'$ is reflection of $I$ across $B'C'$. $K'$ is foot from $D'$ to $E'F'$. This means points $\{K',D'\}$ and $\{I,M'\}$ are symmetric across $B'C'$. So $K'D'M'I$ is isosceles trapezoid which obviously cyclic. Inverting back, we find that $K,D,M$ are colinear.

Invert about $K$ with radius $KI$. Let $Z^*$ denote the image of $Z$ under this inversion. By the incenter-excenter lemma $KI=KB=KC$, so $I,B,C$ are fixed by the inversion. Note that $A\mapsto A^*$, where $A^*=AI\cap BC$. Hence the circle with diameter $AI$ maps to the circle with diameter $A^*I$. Furthermore $(ABC)\mapsto BC$, so $M^*$ is the intersection of $BC$ and the circle with diameter $A^*I$. But this intersection is clearly $D$. Therefore $M^*=D$, which implies $M$ and $D$ are collinear with the center of inversion. Thus, $M,D,K$ are collinear and we are done.

Let $\Delta DEF $ be the intouch triangle WRT $\Delta ABC$ with $I $ as incenter and $M,M'$ as midpoints of arc $BC $ not containing $A $ and arc $BAC $. Let $AI \cap BC $ $=$ $L $. Let $M'I $ $\cap $ $\odot (ABC) $ $=$ $T $. Let $\odot (AFE) $ $\cap $ $\odot (ABC) $ $=$ $K $ Obviously $\angle AKI=90^{\circ} $, so if $KI \cap \odot (ABC) $ $=$ $P $, then, $P $ is $A- $antipode. The line through $I $ perpendicular to $AI $ is ofcourse tangent to $\odot (AFE) $ and $\odot (IBC) $. Also, $T $ is the $A- $mixtilinear incircle touch point. By Radical Axes Theorem, $AK,$ $I-\text {tangent} $ and $BC $ are concurrent, say at $G $. Let $N $ be other tangent from $G $ to $\odot (BIC) $. So, from La Hire's Theorem $\implies $ $N $ $\in $ $M'I $. Since, $MT \perp IN $ $\implies $ $IT=NT $ $\implies $ $G \in TM $ . Also, $GKIT $ is cyclic. $$\angle MKI=\angle MAO=\angle AMO=\angle ITD=\angle DKI $$

Much Shorter: Perform Inversion about $\odot (BIC)$ and note, $K,D$ get swapped $\implies$ $M \in KD$ $\qquad \blacksquare$

Invert about the incircle. Notice $\triangle A'B'C'$ is the medial triangle of $\triangle DEF,$ $K'$ is the foot from $D$ to $\overline{EF},$ and $I$ is the orthocenter of $\triangle A'B'C'.$ Hence, $\overline{K'I}$ is the reflection of $\overline{DM'}$ in $\overline{B'C'}$ so $IM'DK'$ is cyclic. $\square$

parmenides51 wrote: Let $ABC$ be a triangle such that $AB\ne AC$ and $\omega$ be the circumcircle of this triangle. Let $I$ be the center of the inscribed circle of $ABC$ which touches $BC$ at $D$. Let the circle with diameter $AI$ meets $\omega$ again at $K$. If the line $AI$ intersects $\omega$ again at $M$, show that $K, D, M$ are collinear. Without inversion and spiral simirality.We find with angle chansing $\angle$$MID$$=$$2$$×$$\beta$$+$$\alpha$$-90$ $(1)$.$K^{'}$ is intersection of $MD$ and circumcircle of $ABC$.We must show that $K^{'}=K$.Then $\angle$$MK^{'}B$$=$$\angle$$CMK$$=$$\angle$$MBC$ then from triangle similarity $MB^2=MD$$\times$$MK^{'}$ besides we know that $MB=MI=MC$ (well known) then $MI^2=MD$$\times$$MK^{'}$ $(2)$ We get from $(1)$ and $(2)$ $\angle$$MID$$=$$\angle$$IK^{'}D$$=$$2×\beta+\alpha-90$ $(3)$ On the other hand $\angle$$MIA$$=$$\angle$$MBC$$+$$\angle$$CBA$$=$$\alpha$$+$$2$$\beta$ $(4)$ After we get from $(3)$ and $(4)$ $\angle$$IK^{'}A$$=$$90$ then $K^{'}=K$