Prove that the 3 lines are concurrent

Question

Let triangle $ABC$ be such that $AB<AC$, with circumcenter $O$, incenter $I$ ; let $M$ be the midpoint of $BC$, $N$ the midpoint of arc $BC$ of the circumcirle of $ABC$ containing $A$. The circumcircles of triangles $IAN$ and $IBM$ intersect at a point $K$ other than $I$. $BK$ intersects $AC$ at $X$, $NK$ intersects $AI$ at $Y$. Prove that $BI$, $AN$, $XY$ are concurrent.

(Diagram in Geogebra)

I tried but couldn't find a bright direction for this problem, someone help me, give me a suggestion, thank you very much.

I think that if we call D the intersection of AN and BI we can prove that X,Y,D are collinear. Let F be the intersection of AC and BD, then (AF,ID)=-1. We will try Menelaus proof for triangle AIF.

Answer 1

Particular case, where the triangle is isosceles and $AC=BC$:

May this idea helps: As can be seen in figure points C, O and I are co-linear and CR is perpendicular bisector of AB. The radii of small circles are equal and IN and KP are diameters of the top circle with center Q. Angles KPI and KNI are equal, hence $KN||IP$. We connect D to Y, X is where it meets AC. Now we show that the extension of BK meets point X. Since arc IK is common between two equal circles then $\angle KBI=\angle KPI=\angle KNI$ this means that the extension of BK, must be a diagonal of right angled trapezoid KIXY which deduce it meets point X.

Answer 2

Here is a solution using barycentric coordinates. It is given to show after the job is done, as a bonus, other properties that may be useful to give a "simple" geometric solution. Here is a picture of some interesting green points, mentioned with their properties in the aftermath.

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Notations: We use $a,b,c$ for the sides of the triangle. A point $P$ in the plane of the given $\Delta ABC$ has norm(aliz)ed barycentric coordinates $x,y,z$, $x+y+z=1$, if( and only i)f we have the relation $P=xA+yB+zC$, seen as a relation using afixes or vectorially as $ \overrightarrow{\Omega P}= x\overrightarrow{\Omega A} + y\overrightarrow{\Omega B} + z\overrightarrow{\Omega C} $ for one (and thus any) reference point $\Omega$ in the same plane. We then write $P=(x,y,z)$. Often, involved expressions for $x,y,z$ have a heavy denominator, that is not so important. In such cases we may want to write $P=[x:y:z]$ for the point with normed barycentric coordinates $\left(\frac x{x+y+z},\ \frac y{x+y+z},\ \frac z{x+y+z}\right)$. By abuse, i will also write $\frac 1{x+y+z}(x,y,z)$ instead during computations.

We call then $[x:y:z]$ also (not (necessarily) normed) barycentric coordinates. For the most points of the triangle we have know expressions. We can compute now very quickly the nature of the needed concurrence point.

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The solution:

Below, we use $s$ for the semiperimeter, $s=(a+b+c)/2$. Then: $$ \begin{aligned} &&A &=(1,0,0)\ , \\ &&B &=(0,1,0)\ , \\ &&C &=(0,0,1)\ , \\ &&M &=\left(\frac 12,\frac 12,0\right)=\frac 12(1,1,0)=[1:1:0]\ , \\ &&I &=\frac 1{a+b+c}(a,b,c)=[a:b:c]\ , \\ &\text{circle }\odot(ABC)\ :& 0&=a^2yz + b^2zx+c^2xy\ , \\ &\text{line }AI\ :&\ 0&=\begin{vmatrix}x & y & z\\ 1 & 0 & 0 \\ a & b & c\end{vmatrix}\ ,\qquad\text{ i.e. }\qquad\frac yb=\frac zc\ , \\ &E=AI\cap\odot(ABC)& E&=\left[-\frac{a^2}{b+c}\ :\ b:\ c\right]=[-a^2\ :\ b^2 +bc\ :\ c^2 +bc]\ ,\\ &\text{line }EM\ :&\ 0& =\begin{vmatrix} x & y & z\\ -a^2 & b^2+bc & c^2+bc \\ 0 & 1 & 1 \end{vmatrix}=\begin{vmatrix} x & y-z & *\\ -a^2 & b^2-c^2 & * \\ 0 & 0 & 1 \end{vmatrix}\ , \\ &N\in EM\cap\odot(ABC)& N&=[-a^2\ :\ b^2 -bc\ :\ c^2 -bc]\ , \\[3mm] &\text{circle }\odot(AIN)\ :& 0&=-a^2yz - b^2zx-c^2xy +\frac {abc}2(x+y+z)\left(\frac yb+\frac zc\right)\ , \\[3mm] &\text{circle }\odot(BIM)\ :& 0&=-a^2yz - b^2zx-c^2xy +\frac 12(x+y+z)(c(2b-a)x+a^2z)\ , \\ &K\in\odot(AIN)\cap \odot(BIM) & K&=[a^2(s-a)\ :\ (2b-a)^2(s-b)\ :\ c(2b-a)(b-c)]\ , \\ &\text{line }AC\ :&\ 0&=y\ , \\ &\text{line }BK\ :&\ 0&= \begin{vmatrix} x & y & z\\ 0 & 1& 0 \\ a^2(s-a) & * & c(2b-a)(b-c) \end{vmatrix}\ , \\ &X=AC\cap BK & X&=[a^2(s-a) \ :\ 0\ :\ c(2b-a)(b-c)]\ , \\[3mm] &\text{bisector }AI\ :&\ 0&=\frac yb-\frac zc\ , \\ &\text{line }NK\ :&\ 0&= \begin{vmatrix} x & y & z\\ -a^2 & b(b-c) &-c(b-c) \\ a^2(s-a) & (2b-a)^2(s-b) & c(2b-a)(b-c) \end{vmatrix}\ , \\ &Y=AI\cap NK& Y&=\left[a\left(\frac a{b-c}-\frac{2b}{2b-a}\right)\ :\ b\ :\ c\right]\ , \\[3mm] &\text{bisector }BI\ :&\ 0&=\frac xa-\frac zc\ , \\ &\text{outer bisector }AN\ :&\ 0&=\frac yb+\frac zc\ , \\ &I_B=AN\cap BI & I_B&=[a:-b:c]\ . \\[3mm] \end{aligned} $$

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And it remains to check that the points $I_B,X,Y$ are collinear, which is algebraically the vanishing of a determinant. We may use $\sim$ to denote an equality up to a non-zero factor. $$ \begin{aligned} &\begin{vmatrix} a &-b&c\\ a^2(s-a) & 0 & c(2b-a)(b-c)\\ a\left(\frac a{b-c}-\frac{2b}{2b-a}\right) & b & c \end{vmatrix} \sim \begin{vmatrix} 1 &-1&1\\ a(s-a) & 0 & (2b-a)(b-c)\\ \frac a{b-c}-\frac{2b}{2b-a} & 1 & 1 \end{vmatrix} \\ &\qquad \sim \begin{vmatrix} 1 &-1&1\\ a(s-a) & 0 & (2b-a)(b-c)\\ 1+\frac a{b-c}-\frac{2b}{2b-a} & 0 & 2 \end{vmatrix} \sim \begin{vmatrix} a(s-a) & (2b-a)(b-c)\\ \frac a{b-c}-\frac a{2b-a} & 2 \end{vmatrix} \\ &\qquad =a \begin{vmatrix} 2(s-a) & 1\\ (2b-a)-(b-c) &1 \end{vmatrix} =a \begin{vmatrix} b+c-a & 1\\ b+c-a &1 \end{vmatrix}=0\ . \end{aligned} $$ This shows that $XY$ passes through the ex-center $I_B$.

$\square$

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Bonus: The job is done. But let us mention the following properties, that may allow to construct the complicated point $K$ in a different manner, and thus to attack the problem from a different perspective.

Let $S\in BC$ be the point $S=[0:a-b:c]=\frac 1{s-b}(0,a-b,c)$. Then $S$ is on the line $KI$ and on the circle $\odot(NKM)$.

Let $T\in AB$ be the point $T=[a:2b-a:0]=\frac 1{2b}(a,2b-a,0)$. Then $T$ is also on the line $KI$.

Let $U$ be the second point of intersection ($U\ne A$) of the circle $\odot(AIN)$ with the line $AC$. Then $U=(a:0:2b-a)$. In particular, $TU\|BC$, since the points $T,U$ are separating $AB$, $AC$ in the same proportion.

Let $P$ be the mid point of $AB$.
Let $V$ be the point $V=[a:b:a-b]$. Then $V$ is on the parallel $MP$ to $AC$, on the parallel $TU$ to $BC$, and on the angle bisector $CI$.