Subfield: Geometry


Question: Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.

Solution:
Let $S$ be the intersection point of $B C$ and the angle bisector of $\angle B A D$, and let $T$ be the intersection point of $B C$ and the angle bisector of $\angle B X C$. We will prove that both quadruples $A, I, B, S$ and $A, I, B, T$ are concyclic, which yields $S=T$.

Firstly denote by $M$ the middle of arc $A B$ of the circumcenter of $A B C$ which does not contain $C$. Consider the circle centered at $M$ passing through $A, I$ and $B$ (it is well-known that $M A=M I=M B$ ); let it intersect $B C$ at $B$ and $S^{\prime}$. Since $\angle B A C>\angle C B A$ it is easy to check that $S^{\prime}$ lies on side $B C$. Denoting the angles in $A B C$ by $\alpha, \beta, \gamma$ we get

$$
\angle B A D=\angle B A C-\angle D A C=\alpha-\beta .
$$

Moreover since $\angle M B C=\angle M B A+\angle A B C=\frac{\gamma}{2}+\beta$, then

$$
\angle B M S^{\prime}=180^{\circ}-2 \angle M B C=180^{\circ}-\gamma-2 \beta=\alpha-\beta .
$$

It follows that $\angle B A S^{\prime}=2 \angle B M S^{\prime}=2 \angle B A D$ which gives us $S=S^{\prime}$.
<img_3823>

Secondly let $N$ be the middle of arc $B C$ of the circumcenter of $A B C$ which does not contain $A$. From $\angle B A C>\angle C B A$ we conclude that $X$ lies on the arc $A B$ of circumcircle of $A B C$ not containing $C$. Obviously both $A I$ and $X T$ are passing through $N$. Since $\angle N B T=\frac{\alpha}{2}=\angle B X N$ we obtain $\triangle N B T \sim \triangle N X B$, therefore

$$
N T \cdot N X=N B^{2}=N I^{2} .
$$

It follows that $\triangle N T I \sim \triangle N I X$. Keeping in mind that $\angle N B C=\angle N A C=\angle I X A$ we get

$$
\angle T I N=\angle I X N=\angle N X A-\angle I X A=\angle N B A-\angle N B C=\angle T B A .
$$

It means that $A, I, B, T$ are concyclic which ends the proof.
Let $\angle B A C=\alpha, \angle A B C=\beta, \angle B C A=\gamma, \angle A C X=\phi$. Denote by $W_{1}$ and $W_{2}$ the intersections of segment $B C$ with the angle bisectors of $\angle B X C$ and $\angle B A D$ respectively. Then $B W_{1} / W_{1} C=B X / X C$ and $B W_{2} / W_{2} D=B A / A D$. We shall show that $B W_{1}=B W_{2}$.



Since $\angle D A C=\angle C B A$, triangles $A D C$ and $B A C$ are similar and therefore

$$
\frac{D C}{A C}=\frac{A C}{B C}
$$

By the Law of sines

$$
\frac{B W_{2}}{W_{2} D}=\frac{B A}{A D}=\frac{B C}{A C}=\frac{\sin \alpha}{\sin \beta}
$$

Consequently

$$
\begin{gathered}
\frac{B D}{B W_{2}}=\frac{W_{2} D}{B W_{2}}+1=\frac{\sin \beta}{\sin \alpha}+1 \\
\frac{B C}{B W_{2}}=\frac{B C}{B D} \cdot \frac{B D}{B W_{2}}=\frac{1}{1-D C / B C} \cdot \frac{B D}{B W_{2}}=\frac{1}{1-A C^{2} / B C^{2}} \cdot \frac{B D}{B W_{2}}= \\
\frac{\sin ^{2} \alpha}{\sin ^{2} \alpha-\sin ^{2} \beta} \cdot \frac{\sin \beta+\sin \alpha}{\sin \alpha}=\frac{\sin \alpha}{\sin \alpha-\sin \beta} .
\end{gathered}
$$

Note that $A X B C$ is cyclic and so $\angle B X C=\angle B A C=\alpha$. Hence, $\angle X B C=180^{\circ}-$ $\angle B X C-\angle B C X=180^{\circ}-\alpha-\phi$. By the Law of sines for the triangle $B X C$, we have

$$
\begin{gathered}
\frac{B C}{W_{1} B}=\frac{W_{1} C}{W_{1} B}+1=\frac{C X}{B X}+1=\frac{\sin \angle C B X}{\sin \phi}+1= \\
\frac{\sin (\alpha+\phi)}{\sin \phi}+1=\sin \alpha \cot \phi+\cos \alpha+1 .
\end{gathered}
$$

So, it's enough to prove that

$$
\frac{\sin \alpha}{\sin \alpha-\sin \beta}=\sin \alpha \cot \phi+\cos \alpha
$$

Since $A C$ is tangent to the circle $A I X$, we have $\angle A X I=\angle I A C=\alpha / 2$. Moreover $\angle X A I=\angle X A B+\angle B A I=\phi+\alpha / 2$ and $\angle X I A=180^{\circ}-\angle X A I-\angle A X I=180^{\circ}-\alpha-\phi$. Applying the Law of sines again $X A C, X A I, I A C$ we obtain

$$
\begin{gathered}
\frac{A X}{\sin (\alpha+\phi)}=\frac{A I}{\sin \alpha / 2}, \\
\frac{A X}{\sin (\gamma-\phi)}=\frac{A C}{\sin \angle A X C}=\frac{A C}{\sin \beta}, \\
\frac{A I}{\sin \gamma / 2}=\frac{A C}{\sin (\alpha / 2+\gamma / 2)} .
\end{gathered}
$$

Combining the last three equalities we end up with

$$
\begin{aligned}
& \frac{\sin (\gamma-\phi)}{\sin (\alpha+\phi)}=\frac{A I}{A C} \cdot \frac{\sin \beta}{\sin \alpha / 2}=\frac{\sin \beta}{\sin \alpha / 2} \cdot \frac{\sin \gamma / 2}{\sin (\alpha / 2+\gamma / 2)} \\
& \frac{\sin (\gamma-\phi)}{\sin (\alpha+\phi)}=\frac{\sin \gamma \cot \phi-\cos \gamma}{\sin \alpha \cot \phi+\cos \alpha}=\frac{2 \sin \beta / 2 \sin \gamma / 2}{\sin \alpha / 2}
\end{aligned}
$$



$$
\frac{\sin \alpha \sin \gamma \cot \phi-\sin \alpha \cos \gamma}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{2 \sin \beta / 2 \cos \alpha / 2}{\cos \gamma / 2}
$$

Subtracting 1 from both sides yields

$$
\begin{gathered}
\frac{-\sin \alpha \cos \gamma-\sin \gamma \cos \alpha}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{2 \sin \beta / 2 \cos \alpha / 2}{\cos \gamma / 2}-1= \\
\frac{2 \sin \beta / 2 \cos \alpha / 2-\sin (\alpha / 2+\beta / 2)}{\cos \gamma / 2}=\frac{\sin \beta / 2 \cos \alpha / 2-\sin \alpha / 2 \cos \beta / 2}{\cos \gamma / 2}, \\
\frac{-\sin (\alpha+\gamma)}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{\sin (\beta / 2-\alpha / 2)}{\cos \gamma / 2}, \\
\frac{-\sin \beta}{\sin \alpha \cot \phi+\cos \alpha}=2 \sin \gamma / 2 \sin (\beta / 2-\alpha / 2)= \\
2 \cos (\beta / 2+\alpha / 2) \sin (\beta / 2-\alpha / 2)=\sin \beta-\sin \alpha,
\end{gathered}
$$

and the result follows. We are left to note that none of the denominators can vanish.
We first note that

$$
\angle B A D=\angle B A C-\angle D A C=\angle A-\angle B .
$$

Let $C X$ and $A D$ meet at $K$. Then $\angle C X A=\angle A B C=\angle K A C$. Also, we have $\angle I X A=$ $\angle A / 2$, since $\omega$ is tangent to $A C$ at $A$. Therefore,

$$
\angle D A I=|\angle B-\angle A / 2|=|\angle K X A-\angle I X A|=\angle K X I
$$

(the absolute value depends on whether $\angle B \geq \angle A / 2$ or not) which means that $X K I A$ is cyclic, i.e. $K$ lies also on $\omega$.

Let $I K$ meet $B C$ at $E$. (If $\angle B=\angle A / 2$, then $I K$ degenerates to the tangent line to $\omega$ at I.) Note that $B E I A$ is cyclic, because

$$
\angle E I A=180^{\circ}-\angle K X A=180^{\circ}-\angle A B E .
$$

We have $\angle E K A=180^{\circ}-\angle A X I=180^{\circ}-\angle A / 2$ and $\angle A E I=\angle A B I=\angle B / 2$. Hence

$$
\begin{aligned}
\angle E A K & =180^{\circ}-\angle E K A-\angle A E I \\
& =180^{\circ}-\left(180^{\circ}-\angle A / 2\right)-\angle B / 2 \\
& =(\angle A-\angle B) / 2 \\
& =\angle B A D / 2 .
\end{aligned}
$$

This means that $A E$ is the angle bisector of $\angle B A D$. Next, let $M$ be the point of intersection of $A E$ and $B I$. Then

$$
\angle E M I=180^{\circ}-\angle B / 2-\angle B A D / 2=180^{\circ}-\angle A / 2 \text {, }
$$



and so, its supplement is

$$
\angle A M I=\angle A / 2=\angle A X I,
$$

so $X, M, K, I, A$ all lie on $\omega$. Next, we have

$$
\begin{aligned}
\angle X M A & =\angle X K A \\
& =180^{\circ}-\angle A D C-\angle X C B \\
& =180^{\circ}-\angle A-\angle X C B \\
& =\angle B+\angle X C A \\
& =\angle B+\angle X B A \\
& =\angle X B E,
\end{aligned}
$$

and so $X, B, E, M$ are concyclic. Hence

$$
\begin{aligned}
\angle E X C & =\angle E X M+\angle M X C \\
& =\angle M B E+\angle M A K \\
& =\angle B / 2+\angle B A D / 2 \\
& =\angle A / 2 \\
& =\angle B X C / 2 .
\end{aligned}
$$

This means that $X E$ is the angle bisector of $\angle B X C$ and so we are done!



<img_3517>
It is $\angle A B D=\angle D A C$, and so $\overline{A C}$ is tangent to the circumcircle of $\triangle B A D$ at $A$. Hence $C A^{2}=C D \cdot C B$.



<img_3716>

Triangle $\triangle A B C$ is similar to triangle $\triangle C A D$, because $\angle C$ is a common angle and $\angle C A D=\angle A B C$, and so $\angle A D C=\angle B A C=2 \varphi$.

Let $Q$ be the point of intersection of $\overline{A D}$ and $\overline{C X}$. Since $\angle B X C=\angle B A C=2 \varphi$, it follows that $B D Q X$ is cyclic. Therefore, $C D \cdot C B=C Q \cdot C X=C A^{2}$ which implies that $Q$ lies on $\omega$.

Next let $P$ be the point of intersection of $\overline{A D}$ with the circumcircle of triangle $\triangle A B C$. Then $\angle P B C=\angle P A C=\angle A B C=\angle A P C$ yielding $C A=C P$. So, let $T$ be on the side $\overline{B C}$ such that $C T=C A=C P$. Then

$$
\angle T A D=\angle T A C-\angle D A C=\left(90^{\circ}-\frac{\angle C}{2}\right)-\angle B=\frac{\angle A-\angle B}{2}=\frac{\angle B A D}{2}
$$

that is, line $\overline{A T}$ is the angle bisector of $\angle B A D$. We want to show that $\overline{X T}$ is the angle bisector of $\angle B X C$. To this end, it suffices to show that $\angle T X C=\varphi$.

It is $C T^{2}=C A^{2}=C Q \cdot C X$, and so $\overline{C T}$ is tangent to the circumcircle of $\triangle X T Q$ at $T$. Since $\angle T X Q=\angle Q T C$ and $\angle Q D C=2 \varphi$, it suffices to show that $\angle T Q D=\varphi$, or, in other words, that $I, Q$, and $T$ are collinear.

Let $T^{\prime}$ is the point of intersection of $\overline{I Q}$ and $\overline{B C}$. Then $\triangle A I C$ is congruent to $\triangle T^{\prime} I C$, since they share $\overline{C I}$ as a common side, $\angle A C I=\angle T^{\prime} C I$, and

$$
\angle I T^{\prime} D=2 \varphi-\angle T^{\prime} Q D=2 \varphi-\angle I Q A=2 \varphi-\angle I X A=\varphi=\angle I A C .
$$

Therefore, $C T^{\prime}=C A=C T$, which means that $T$ coincides with $T^{\prime}$ and completes the proof.
Let $G$ be the point of intersection of $\overline{A D}$ and $\overline{C X}$. Since the quadrilateral $A X B C$ is cyclic, it is $\angle A X C=\angle A B C$.



<img_3355>

Let the line $\overline{A D}$ meet $\omega$ at $K$. Then it is $\angle A X K=\angle C A D=\angle A B C$, because the angle that is formed by a chord and a tangent to the circle at an endpoint of the chord equals the inscribed angle to that chord. Therefore, $\angle A X K=\angle A X C=\angle A X G$. This means that the point $G$ coincides with the point $K$ and so $G$ belongs to the circle $\omega$.

Let $E$ be the point of intersection of the angle bisector of $\angle D A B$ with $\overline{B C}$. It suffices to show that

$$
\frac{C E}{B E}=\frac{X C}{X B}
$$

Let $F$ be the second point of intersection of $\omega$ with $\overline{A B}$. Then we have $\angle I A F=\frac{\angle C A B}{2}=$ $\angle I X F$, where $I$ is the incenter of $\triangle A B C$, because $\angle I A F$ and $\angle I X F$ are inscribed in the same arc of $\omega$. Thus $\triangle A I F$ is isosceles with $A I=I F$. Since $I$ is the incenter of $\triangle A B C$, we have $A F=2(s-a)$, where $s=(a+b+c) / 2$ is the semiperimeter of $\triangle A B C$. Also, it is $C E=A C=b$ because in triangle $\triangle A C E$, we have

$$
\begin{aligned}
\angle A E C & =\angle A B C+\angle B A E \\
& =\angle A B C+\frac{\angle B A D}{2} \\
& =\angle A B C+\frac{\angle B A C-\angle A B C}{2} \\
& =90^{\circ}-\frac{\angle A C E}{2},
\end{aligned}
$$

and so $\angle C A E=180^{\circ}-\angle A E C-\angle A C E=90^{\circ}-\frac{\angle A C E}{2}=\angle A E C$. Hence

$$
B F=B A-A F=c-2(s-a)=a-b=C B-C E=B E \text {. }
$$

Moreover, triangle $\triangle C A X$ is similar to triangle $\triangle B F X$, because $\angle A C X=\angle F B X$ and

$$
\angle X F B=\angle X A F+\angle A X F=\angle X A F+\angle C A F=\angle C A X .
$$

Therefore

$$
\frac{C E}{B E}=\frac{A C}{B F}=\frac{X C}{X B}
$$

as desired. The proof is complete.
Let $\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $Y$ be the second point of intersection of the circle $\omega$ with the line $A D$. Let $L$ be the intersection of $B C$ with the angle bisector of $\angle B A D$. We will prove $\angle L X C=$ $1 / 2 \angle B A C=1 / 2 \angle B X C$.

We will refer to the angles of $\triangle A B C$ as $\angle A, \angle B, \angle C$. Thus $\angle B A D=\angle A-\angle B$.

On the circumcircle of $\triangle A B C$, we have $\angle A X C=\angle A B C=\angle C A D$, and since $A C$ is tangent to $\omega$, we have $\angle C A D=\angle C A Y=\angle A X Y$. Hence $C, X, Y$ are collinear.

Also note that $\triangle C A L$ is isosceles with $\angle C A L=\angle C L A=\frac{1}{2}(\angle B A D)+\angle A B C=\frac{1}{2}(\angle A+$ $\angle B)$ hence $A C=C L$. Moreover, $C I$ is angle bisector to $\angle A C L$ so it's the symmetry axis for the triangle, hence $\angle I L C=\angle I A C=1 / 2 \angle A$ and $\angle A L I=\angle L I A=1 / 2 \angle B$. Since $A C$ is tangent to $\omega$, we have $\angle A Y I=\angle I A C=1 / 2 \angle A=\angle L A Y+\angle A L I$. Hence $L, Y, I$ are collinear.

Since $A C$ is tangent to $\omega$, we have $\triangle C A Y \sim \triangle C X A$ hence $C A^{2}=C X \cdot C Y$. However we proved $C A=C L$ hence $C L^{2}=C X \cdot C Y$. Hence $\triangle C L Y \sim \triangle C X L$ and hence $\angle C X L=\angle C L Y=\angle C A I=1 / 2 \angle A$.

<img_3734>
Let $M$ be the midpoint of the arc $B C$. Let $\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $N$ be the second point of intersection of $\omega$ with $A B$ and $L$ the intersection of $B C$ with the angle bisector of $\angle B A D$. We know $\frac{D L}{L B}=\frac{A D}{A B}$ and want to prove $\frac{X B}{X C}=\frac{L B}{L C}$.

First note that $\triangle C A L$ is isosceles with $\angle C A L=\angle C L A=\frac{1}{2}(\angle B A D)+\angle A B C$ hence $A C=C L$ and $\frac{L B}{L C}=\frac{L B}{A C}$.

Now we calculate $\frac{X B}{X C}$ :

Comparing angles on the circles $\omega$ and the circumcircle of $\triangle A B C$ we get $\triangle X I N \sim$ $\triangle X M B$ and hence also $\triangle X I M \sim \triangle X N B$ (having equal angles at $X$ and proportional adjoint sides). Hence $\frac{X B}{X M}=\frac{N B}{I M}$.

Also comparing angles on the circles $\omega$ and the circumcircle of $\triangle A B C$ and using the tangent $A C$ we get $\triangle X A I \sim \triangle X C M$ and hence also $\triangle X A C \sim \triangle X I M$. Hence $\frac{X C}{X M}=$ $\frac{A C}{I M}$.

Comparing the last two equations we get $\frac{X B}{X C}=\frac{N B}{A C}$. Comparing with $\frac{L B}{L C}=\frac{L B}{A C}$, it remains to prove $N B=L B$.



<img_3549>

We prove $\triangle I N B \equiv \triangle I L B$ as follows:

First, we note that $I$ is the circumcentre of $\triangle A L N$. Indeed, $C I$ is angle bisector in the isosceles triangle $A C L$ so it's perpendicular bisector for $A L$. As well, $\triangle I A N$ is isosceles with $\angle I N A=\angle C A I=\angle I A B$ hence $I$ is also on the perpendicular bisector of $A N$.

Hence $I N=I L$ and also $\angle N I L=2 \angle N A L=\angle A-\angle B=2 \angle N I B$ (the last angle is calculated using that the exterior angle of $\triangle N I B$ is $\angle I N A=\angle A / 2$. Hence $\angle N I B=$ $\angle L I B$ and $\triangle I N B \equiv \triangle I L B$ by SAS.
Let $M, N$ be the midpoints of arcs $B C, B A$ of the circumcircle $A B C$, respectively. Let $Y$ be the second intersection of $A D$ and circle $A B C$. Let $E$ be the incenter of triangle $A B Y$ and note that $E$ lies on the angle bisectors of the triangle, which are the lines $Y N$ (immediate), $B C$ (since $\angle C B Y=\angle C A Y=\angle C A D=\angle A B C)$ and the angle bisector of $\angle D A B$; so the question reduces to showing that $E$ is also on $X M$, which is the angle bisector of $\angle C X B$.

We claim that the three lines $C X, A D Y, I E$ are concurrent at a point $D^{\prime}$. We will complete the proof using this fact, and the proof will appear at the end (and see the solution by HEL5 for an alternative proof of this fact).

To show that $X E M$ are collinear, we construct a projective transformation which projects $M$ to $X$ through center $E$. We produce it as a composition of three other projections. Let $O$ be the intersection of lines $A D^{\prime} D Y$ and $C I N$. Projecting the points $Y N C M$ on the circle $A B C$ through the (concyclic) point $A$ to the line $C N$ yields the points $O N C I$. Projecting these points through $E$ to the line $A Y$ yields $O Y D D^{\prime}$ (here we use the facts that $D^{\prime}$ lies on $I E$ and $A Y$ ). Projecting these points to the circle $A B C$ through $C$ yields $N Y B X$ (here we use the fact that $D^{\prime}$ lies on $C X$ ). Composing, we observe that we found a projection of the circle $A B C$ to itself sending $Y N C M$ to $N Y B X$. Since the projection of the circle through $E$ also sends $Y N C$ to $N Y B$, and three points determine a projective transformation, the projection through $E$ also sends $M$ to $X$, as claimed.



<img_3265>

Let $B^{\prime}, D^{\prime}$ be the intersections of $A B, A D$ with the circle $A X I$, respectively. We wish to show that this $D^{\prime}$ is the concurrency point defined above, i.e. that $C D^{\prime} X$ and $I D^{\prime} E$ are collinear. Additionally, we will show that $I$ is the circumcenter of $A B^{\prime} E$.

Consider the inversion with center $C$ and radius $C A$. The circles $A X I$ and $A B D$ are tangent to $C A$ at $A$ (the former by definition, the latter since $\angle C A D=\angle A B C$ ), so they are preserved under the inversion. In particular, the inversion transposes $D$ and $B$ and preserves $A$, so sends the circle $C A B$ to the line $A D$. Thus $X$, which is the second intersection of circles $A B C$ and $A X I$, is sent by the inversion to the second intersection of $A D$ and circle $A X I$, which is $D^{\prime}$. In particular $C D^{\prime} X$ are collinear.

In the circle $A I B^{\prime}, A I$ is the angle bisector of $B^{\prime} A$ and the tangent at $A$, so $I$ is the midpoint of the $\operatorname{arc} A B^{\prime}$, and in particular $A I=I B^{\prime}$. By angle chasing, we find that $A C E$ is an isosceles triangle:

$\angle C A E=\angle C A D+\angle D A E=\angle A B C+\angle E A B=\angle A B E+\angle E A B=\angle A E B=\angle A E C$,

thus the angle bisector $C I$ is the perpendicular bisector of $A E$ and $A I=I E$. Thus $I$ is the circumcenter of $A B^{\prime} E$.

We can now show that $I D^{\prime} E$ are collinear by angle chasing:

$$
\angle E I B^{\prime}=2 \angle E A B^{\prime}=2 \angle E A B=\angle D A B=\angle D^{\prime} A B^{\prime}=\angle D^{\prime} I B^{\prime} .
$$
Let $W$ be the midpoint of $\operatorname{arc} B C$, let $D^{\prime}$ be the second intersection point of $A D$ and the circle $A B C$. Let $P$ be the intersection of the angle bisector $X W$ of $\angle C X B$ with $B C$; we wish to prove that $A P$ is the angle bisector of $D A B$. Denote $\alpha=\frac{\angle C A B}{2}, \beta=\angle A B C$.

Let $M$ be the intersection of $A D$ and $X C$. Angle chasing finds:

$$
\begin{aligned}
\angle M X I & =\angle A X I-\angle A X M=\angle C A I-\angle A X C=\angle C A I-\angle A B C=\alpha-\beta \\
& =\angle C A I-\angle C A D=\angle D A I=\angle M A I
\end{aligned}
$$

And in particular $M$ is on $\omega$. By angle chasing we find

$$
\angle X I A=\angle I X A+\angle X A I=\angle I C A+\angle X A I=\angle X A C=\angle X B C=\angle X B P
$$

and $\angle P X B=\alpha=\angle C A I=\angle A X I$, and it follows that $\triangle X I A \sim \triangle X B P$. Let $S$ be the second intersection point of the cirumcircles of $X I A$ and $X B P$. Then by the spiral map lemma (or by the equivalent angle chasing) it follows that $I S B$ and $A S P$ are collinear.



Let $L$ be the second intersection of $\omega$ and $A B$. We want to prove that $A S P$ is the angle bisector of $\angle D A B=\angle M A L$, i.e. that $S$ is the midpoint of the $\operatorname{arc} M L$ of $\omega$. And this follows easily from chasing angular arc lengths in $\omega$ :

$$
\begin{aligned}
& \overparen{A I}=\angle C A I=\alpha \\
& \overparen{I L}=\angle I A L=\alpha \\
& \overparen{M I}=\angle M X I=\alpha-\beta \\
& \overparen{A I}-\overparen{S L}=\angle A B I=\frac{\beta}{2}
\end{aligned}
$$

And thus

$$
\overparen{M L}=\overparen{M I}+\overparen{I L}=2 \alpha-\beta=2\left(\overparen{A I}-\frac{\beta}{2}\right)=2 \overparen{S L}
$$

<img_3694>
Let $E$ be the intersection of the bisector of $\angle B A D$ and $B C$, and $N$ be the middle point of arc $B C$ of the circumcircle of $A B C$. Then it suffices to show that $E$ is on line $X N$.

We consider the inversion at $A$. Let $P^{*}$ be the image of a point denoted by $P$. Then $A, B^{*}, C^{*}, E^{*}$ are concyclic, $X^{*}, B^{*}, C^{*}$ are colinear, and $X^{*} I^{*}$ and $A C^{*}$ are parallel. Now it suffices to show that $A, X^{*}, E^{*}, N^{*}$ are concyclic. Let $Y$ be the intersection of $B^{*} C^{*}$ and $A E^{*}$. Then, by the power of a point, we get

$$
\begin{aligned}
A, X^{*}, E^{*}, N^{*} \text { are concyclic } & \Longleftrightarrow Y X^{*} \cdot Y N^{*}=Y A \cdot Y E^{*} \\
& \Longleftrightarrow Y X^{*} \cdot Y N^{*}=Y B^{*} \cdot Y C^{*} \\
& \left(A, B^{*}, C^{*}, E^{*} \text { are concyclic }\right)
\end{aligned}
$$

Here, by the property of inversion, we have

$$
\angle A I^{*} B^{*}=\angle A B I=\frac{1}{2} \angle A B C=\frac{1}{2} \angle C^{*} A D^{*} \text {. }
$$



<img_3856>

Define $Q, R$ as described in the figure, and we get by simple angle chasing

$$
\angle Q A I^{*}=\angle Q I^{*} A, \quad \angle R A I^{*}=\angle B^{*} I^{*} A
$$

Especially, $B^{*} R$ and $A I^{*}$ are parallel, so that we have

$$
\frac{Y B^{*}}{Y N^{*}}=\frac{Y R}{Y A}=\frac{Y X^{*}}{Y C^{*}}
$$

and the proof is completed.