Dadgarnia wrote: Consider a triangle $ABC$ with incenter $I$. Let $D$ be the intersection of $BI,AC$ and $CI$ intersects the circumcircle of $ABC$ at $M$. Point $K$ lies on the line $MD$ and $\angle KIA=90^\circ$. Let $F$ be the reflection of $B$ about $C$. Prove that $BIKF$ is cyclic. Firstly, observe that it suffices to show $\angle KFB + \angle KIB = 180^{\circ} \iff \angle KFB = \frac{\angle C}{2} \iff \overline{FK} \parallel \overline{IC} $. Consider a homothety centered at $B$ with a dilation factor $\frac{1}{2}$. So, now it suffices to show that $IC$ bisects $BK$. We proceed with barycentric coordinates with respect to $\triangle ABC$ and consider the usual setup. The major task is to evaluate the coordinates of $K$. To do this, we intersect lines $MD$ and line through $I$ tangent to $\odot(BIC)$. Let $T,J$ be points on $AB, AC$ respectively so that $AT = AJ$ and $I$ is the mid-point of $TJ.$ (These are the same points that appear in the mixtilinear configuration).

Trivial Coordinates

Equations of Lines

The point K

Finally, obtain the mid-point of $KB$ and show that it homogenizes to the form $(a:b:x)$ for some $x$.

Here is synthetic solution. I'm sure everything can be written just in terms of harmonic bundles, but I don't see how Let $B_1$ be the reflection of $B$ in $I$, $E$ be the foot of the exterior $B$-bisector, $F$ be the intersection of the parallel in $B$ to $IC$ with line $ME$ and take $S=MD\cap AI,X=MI\cap BS,Y=AB\cap MI$, $N$ the midpoint of $\overline{AI}$ and $I_a,I_b,I_c$ the excenters. Step 1.$ F\in AI$ Proof: $$(M,IC_{\infty}; I_c,I)=-1=(E,B;I_c,I_a)\stackrel{F}{=} (M,IC_{\infty}; I_c,FI_a\cap IC)$$so $I\in FI_a$ which gives $F\in AI.$ Step 2. $NX\parallel AB$ Proof: $$(I,Y;X,IC_{\infty})\stackrel{B}{=} (I,A; S,F)\stackrel{M}{=} (C,A; D,E)=-1$$so $X$ is the midpoint os $\overline{IY}$. Hence $\overline{NX}$ is the $I$-middle line in $\triangle{AIY}$, hence $NX\parallel AB.$ Consequence of Step 2: $$\frac{SN}{SA}=\frac{SX}{BX}\stackrel{\text{Menelaus}}{=} \frac{DI}{BI}\cdot \frac{MS}{DM}$$and so $$\frac{DI}{IB_1}=\frac{NS}{NI}\cdot\frac{DM}{MS}.$$ Note that $MN\parallel IK$ yields $\frac{NS}{NI}=\frac{MD}{MK}$, so the above equality can be rewritten as $\frac{DI}{IB_1}=\frac{MD}{MK}$, meaning that $KB_1\parallel MI$ i.e. $KF\parallel IC.$ This is easily seen to imply the problem.

Dear Mathlinkers, a proof using https://artofproblemsolving.com/community/c6h1898293_parallel_to_a_side https://artofproblemsolving.com/community/c6h1898384_two_parallels https://artofproblemsolving.com/community/c6t48f6h1898786_two_parallels and finishing with the Reim's theorem... Sincerely Jean-Louis

DK-(O)=E by Reim BIEK cyclic, MK/MD=NI/ND=NA/ND=CB/CD=IB/ID so MI bisect BK thus FK//CI so BFK=BCI=BED so BEKF cyclic so BIKF cylic

Let $MD \cap \odot (ABC) = $ {${M,P}$}. $\angle BIK = 180^{\circ}-\frac{C}{2}$ and $\angle BPK = \frac{C}{2}$. $\therefore BIKP$ is cyclic. So it suffices to prove that $BIPF$ is cyclic. Lets perform $\sqrt{ac}$ inversion with centre $B$ accompanied by reflection in $\angle$ bisector of $\angle ABC$. Under this transformation, $A$ and $C$ are swapped, $I$ goes to $I_{B}$ (excentre of $\triangle ABC$ opp. to B), $D$ goes to $D'$ (midpoint of arc $AC$ not containing $A$), $F$ goes to $F'$ (midpoint of $AB$), $M$ goes to $M'$ ($A-C-M'$ and $BC = CM'$). Lemma 1 : Let $AC \cap F'I_{B} = Q$. Then $QI \parallel AB$. Proof : Apply Menelaus to $\triangle I_{B}BF'$ with $AC$ as transversal. Then $\frac{I_{B}Q}{QF'} = 2\frac{I_{B}D}{DB} = \frac{I_{B}I}{IB}$. This proves the lemma. Lemma 2: $BM'D'Q$ is cyclic. Proof : $QI \parallel AB$. $\therefore \angle CQI = A = \angle CD'I$. So $QICD'$ is cyclic. So $\angle QD'B = \angle QD'I = \angle QCI = \frac{C}{2} = \angle QM'B$. Thus $BM'D'Q$ is cyclic. Lemma 3 : Let $P$ go to $P'$ under this transformation. Then $P' = Q$. Proof : $P$ lies on $\odot (ABC)$ and $MD$. $\therefore P'$ lies on $AC$ and on $\odot BM'D'$. So the lemma is equivalent to showing that $BM'D'Q$ is cyclic, which follows from Lemma 2. Now finally $AC \cap F'I_{B} = P'$. So $F' , I_{B} , P'$ are collinear. So before the transformation (inversion + reflection), $B$, $P$, $I$ and $F$ must be lying on a circle. Thus $BIPF$ is cyclic and this completes the proof. Q.E.D.

Suppose $E = MK \cap (ABC)$, then since $\angle BEM = \angle DIK$, points $B,I,K,E$ are concyclic. Suppose $BC$ intersects $(BIKE)$ at another point $F$, and let $N = BD \cap (ABC)$, $G = CM \cap BK$. Notice that $$\frac{BI}{ID} = \frac{BC}{CD} = \frac{AN}{ND} = \frac{IN}{ND} = \frac{KM}{MD},$$so $S_{\triangle MIB} = S_{\triangle MIK} \Longrightarrow BG=GK$. Finally, by Reim's $KF\parallel IC$, so $BC = CF$ as desired.

Note that $\angle KID = \angle 90 - \angle AID = \frac{\angle C}{2}$ so we need to prove $\angle KFB = \angle ICB$ or $KF || CI$. Let $BK$ meet $CI$ at $S$ and $BI$ meet $ABC$ at $R$. we have $\frac{BS}{SK}.\frac{KM}{DM}.\frac{DI}{IB} = 1$ so we need to prove $\frac{KM}{MD} = \frac{IB}{ID}$. Note that $\frac{IB}{ID} = \frac{BC}{CD} = \frac{RA}{DR} = \frac{RI}{DR}$ so we want to prove $\frac{KD}{DM} = \frac{ID}{RI}$ which is true so $BS = SK$ which implies that $CI || KF$ as wanted.

Let $BI \cap (ABC)=G$, by I-E lemma we have that $MG \perp AI$ which means $MG \parallel IK$ and by ratios $$\frac{BI}{ID}=\frac{BC}{CD}=\frac{GA}{GD}=\frac{GI}{GD}=\frac{MK}{MD} \implies \frac{BI}{MK}=\frac{ID}{MD}=\frac{\sin(\angle KMC)}{\sin(90+\angle BCI)}$$Now by Law of sines and using Ratio Lemma $$\frac{MB}{MK}=\frac{BI \cdot \sin(90-\angle BCI)}{MK \cdot \sin(\angle BAC)}=\frac{\sin(\angle KMC)}{\sin(\angle BAC)} \implies MI \; \text{bisects} \; BK \implies CI \parallel FK$$Now note that $\angle DIK=90-\angle AID=90-(90-\angle BCI)=\angle BCI$ and $\angle BFK=\angle BCI$ so $BIKF$ cyclic as desired thus we are done