EGMO

Solution:

Let $S$ be the intersection point of $BC$ and the angle bisector of $\angle BAD$, and let $T$ be the intersection point of $BC$ and the angle bisector of $\angle BXC$. 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 $\operatorname{arc} AB$ of the circumcenter of $ABC$ which does not contain $C$. Consider the circle centered at $M$ passing through $A, I$ and $B$ (it is well-known that $MA = MI = MB$); let it intersect $BC$ at $B$ and $S'$. Since $\angle BAC > \angle CBA$ it is easy to check that $S'$ lies on side $BC$. Denoting the angles in $ABC$ by $\alpha, \beta, \gamma$ we get
$$\angle BAD = \angle BAC - \angle DAC = \alpha - \beta.$$
Moreover since $\angle MBC = \angle MBA + \angle ABC = \frac{\gamma}{2} + \beta$, then
$$\angle BMS' = 180^\circ - 2\angle MBC = 180^\circ - \gamma - 2\beta = \alpha - \beta$$
It follows that $\angle BAS' = 2\angle BMS' = 2\angle BAD$ which gives us $S = S'$.

Secondly let $N$ be the middle of arc $BC$ of the circumcenter of $ABC$ which does not contain $A$. From $\angle BAC > \angle CBA$ we conclude that $X$ lies on the $\operatorname{arc} AB$ of circumcircle of $ABC$ not containing $C$. Obviously both $AI$ and $XT$ are passing through $N$. Since $\angle NBT = \frac{\alpha}{2} = \angle BXN$ we obtain $\triangle NBT \sim \triangle NXB$, therefore
$$NT \cdot NX = NB^2 = NI^2$$
It follows that $\triangle NTI \sim \triangle NIX$. Keeping in mind that $\angle NBC = \angle NAC = \angle IXA$ we get
$$\angle TIN = \angle IXN = \angle NXA - \angle IXA = \angle NBA - \angle NBC = \angle TBA.$$
It means that $A, I, B, T$ are concyclic which ends the proof.

Solution:

Let $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle BCA = \gamma$, $\angle ACX = \phi$. Denote by $W_1$ and $W_2$ the intersections of segment $BC$ with the angle bisectors of $\angle BXC$ and $\angle BAD$ respectively. Then $BW_1 / W_1C = BX / XC$ and $BW_2 / W_2D = BA / AD$. We shall show that $BW_1 = BW_2$.

Since $\angle DAC = \angle CBA$, triangles $ADC$ and $BAC$ are similar and therefore
$$\frac{DC}{AC} = \frac{AC}{BC}$$
By the Law of sines
$$\frac{BW_2}{W_2D} = \frac{BA}{AD} = \frac{BC}{AC} = \frac{\sin \alpha}{\sin \beta}.$$
Consequently
$$\begin{gathered} \frac{BD}{BW_2} = \frac{W_2D}{BW_2} + 1 = \frac{\sin \beta}{\sin \alpha} + 1, \\ \frac{BC}{BW_2} = \frac{BC}{BD} \cdot \frac{BD}{BW_2} = \frac{1}{1 - DC/BC} \cdot \frac{BD}{BW_2} = \frac{1}{1 - AC^2 / BC^2} \cdot \frac{BD}{BW_2} = \\ \frac{\sin^2 \alpha}{\sin^2 \alpha - \sin^2 \beta} \cdot \frac{\sin \beta + \sin \alpha}{\sin \alpha} = \frac{\sin \alpha}{\sin \alpha - \sin \beta}. \end{gathered}$$
Note that $AXBC$ is cyclic and so $\angle BXC = \angle BAC = \alpha$. Hence, $\angle XBC = 180^\circ - \angle BXC - \angle BCX = 180^\circ - \alpha - \phi$. By the Law of sines for the triangle $BXC$, we have
$$\begin{gathered} \frac{BC}{W_1B} = \frac{W_1C}{W_1B} + 1 = \frac{CX}{BX} + 1 = \frac{\sin \angle CBX}{\sin \phi} + 1 = \\ \frac{\sin (\alpha + \phi)}{\sin \phi} + 1 = \sin \alpha \cot \phi + \cos \alpha + 1. \end{gathered}$$
So, it's enough to prove that
$$\frac{\sin \alpha}{\sin \alpha - \sin \beta} = \sin \alpha \cot \phi + \cos \alpha$$
Since $AC$ is tangent to the circle $AIX$, we have $\angle AXI = \angle IAC = \alpha / 2$. Moreover $\angle XAI = \angle XAB + \angle BAI = \phi + \alpha / 2$ and $\angle XIA = 180^\circ - \angle XAI - \angle AXI = 180^\circ - \alpha - \phi$. Applying the Law of sines again $XAC, XAI, IAC$ we obtain
$$\begin{gathered} \frac{AX}{\sin (\alpha + \phi)} = \frac{AI}{\sin \alpha / 2}, \\ \frac{AX}{\sin (\gamma - \phi)} = \frac{AC}{\sin \angle AXC} = \frac{AC}{\sin \beta}, \\ \frac{AI}{\sin \gamma / 2} = \frac{AC}{\sin (\alpha / 2 + \gamma / 2)}. \end{gathered}$$
Combining the last three equalities we end up with
$$\begin{gathered} \frac{\sin (\gamma - \phi)}{\sin (\alpha + \phi)} = \frac{AI}{AC} \cdot \frac{\sin \beta}{\sin \alpha / 2} = \frac{\sin \beta}{\sin \alpha / 2} \cdot \frac{\sin \gamma / 2}{\sin (\alpha / 2 + \gamma / 2)} \\ \frac{\sin (\gamma - \phi)}{\sin (\alpha + \phi)} = \frac{\sin \gamma \cot \phi - \cos \gamma}{\sin \alpha \cot \phi + \cos \alpha} = \frac{2 \sin \beta / 2 \sin \gamma / 2}{\sin \alpha / 2} \end{gathered}$$
$$\frac{\sin \alpha \sin \gamma \cot \phi - \sin \alpha \cos \gamma}{\sin \gamma \sin \alpha \cot \phi + \sin \gamma \cos \alpha} = \frac{2 \sin \beta / 2 \cos \alpha / 2}{\cos \gamma / 2}$$
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.

Solution:

We first note that
$$\angle BAD = \angle BAC - \angle DAC = \angle A - \angle B.$$
Let $CX$ and $AD$ meet at $K$. Then $\angle CXA = \angle ABC = \angle KAC$. Also, we have $\angle IXA = \angle A/2$, since $\omega$ is tangent to $AC$ at $A$. Therefore,
$$\angle DAI = |\angle B - \angle A/2| = |\angle KXA - \angle IXA| = \angle KXI,$$
(the absolute value depends on whether $\angle B \geq \angle A/2$ or not) which means that $XKIA$ is cyclic, i.e. $K$ lies also on $\omega$.
Let $IK$ meet $BC$ at $E$. (If $\angle B = \angle A/2$, then $IK$ degenerates to the tangent line to $\omega$ at $I$.) Note that $BEIA$ is cyclic, because
$$\angle EIA = 180^\circ - \angle KXA = 180^\circ - \angle ABE.$$
We have $\angle EKA = 180^\circ - \angle AXI = 180^\circ - \angle A/2$ and $\angle AEI = \angle ABI = \angle B/2$. Hence
$$\begin{aligned} \angle EAK & = 180^\circ - \angle EKA - \angle AEI \\ & = 180^\circ - (180^\circ - \angle A/2) - \angle B/2 \\ & = (\angle A - \angle B)/2 \\ & = \angle BAD/2. \end{aligned}$$
This means that $AE$ is the angle bisector of $\angle BAD$. Next, let $M$ be the point of intersection of $AE$ and $BI$. Then
$$\angle EMI = 180^\circ - \angle B/2 - \angle BAD/2 = 180^\circ - \angle A/2$$
and so, its supplement is
$$\angle AMI = \angle A/2 = \angle AXI$$
so $X, M, K, I, A$ all lie on $\omega$. Next, we have
$$\begin{aligned} \angle XMA & = \angle XKA \\ & = 180^\circ - \angle ADC - \angle XCB \\ & = 180^\circ - \angle A - \angle XCB \\ & = \angle B + \angle XCA \\ & = \angle B + \angle XBA \\ & = \angle XBE, \end{aligned}$$
and so $X, B, E, M$ are concyclic. Hence
$$\begin{aligned} \angle EXC & = \angle EXM + \angle MXC \\ & = \angle MBE + \angle MAK \\ & = \angle B/2 + \angle BAD/2 \\ & = \angle A/2 \\ & = \angle BXC/2 \end{aligned}$$
This means that $XE$ is the angle bisector of $\angle BXC$ and so we are done!

Solution:

It is $\angle ABD = \angle DAC$, and so $\overline{AC}$ is tangent to the circumcircle of $\triangle BAD$ at $A$. Hence $CA^2 = CD \cdot CB$.

Triangle $ABC$ is similar to triangle $CAD$, because $\angle C$ is a common angle and $\angle CAD = \angle ABC$, and so $\angle ADC = \angle BAC = 2\varphi$.
Let $Q$ be the point of intersection of $AD$ and $CX$. Since $\angle BXC = \angle BAC = 2\varphi$, it follows that $BDQX$ is cyclic. Therefore, $CD \cdot CB = CQ \cdot CX = CA^2$ which implies that $Q$ lies on $\omega$.
Next let $P$ be the point of intersection of $AD$ with the circumcircle of triangle $ABC$. Then $\angle PBC = \angle PAC = \angle ABC = \angle APC$ yielding $CA = CP$. So, let $T$ be on the side $BC$ such that $CT = CA = CP$. Then
$$\angle TAD = \angle TAC - \angle DAC = \left(90^\circ - \frac{\angle C}{2}\right) - \angle B = \frac{\angle A - \angle B}{2} = \frac{\angle BAD}{2}$$
that is, line $AT$ is the angle bisector of $\angle BAD$. We want to show that $XT$ is the angle bisector of $\angle BXC$. To this end, it suffices to show that $\angle TXC = \varphi$.
It is $CT^2 = CA^2 = CQ \cdot CX$, and so $CT$ is tangent to the circumcircle of $XTQ$ at $T$. Since $\angle TXQ = \angle QTC$ and $\angle QDC = 2\varphi$, it suffices to show that $\angle TQD = \varphi$, or, in other words, that $I, Q$, and $T$ are collinear.
Let $T'$ be the point of intersection of $IQ$ and $BC$. Then $\triangle AIC$ is congruent to $\triangle T'IC$, since they share $CI$ as a common side, $\angle ACI = \angle T'CI$, and
$$\angle IT'D = 2\varphi - \angle T'QD = 2\varphi - \angle IQA = 2\varphi - \angle IXA = \varphi = \angle IAC.$$
Therefore, $CT' = CA = CT$, which means that $T$ coincides with $T'$ and completes the proof.

Solution:

Let $G$ be the point of intersection of $AD$ and $CX$. Since the quadrilateral $AXBC$ is cyclic, it is $\angle AXC = \angle ABC$.

Let the line $AD$ meet $\omega$ at $K$. Then it is $\angle AXK = \angle CAD = \angle ABC$, 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, $\angle AXK = \angle AXC = \angle AXG$. 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 $\angle DAB$ with $BC$. It suffices to show that
$$\frac{CE}{BE} = \frac{XC}{XB}.$$
Let $F$ be the second point of intersection of $\omega$ with $AB$. Then we have $\angle IAF = \frac{\angle CAB}{2} = \angle IXF$, where $I$ is the incenter of $ABC$, because $\angle IAF$ and $\angle IXF$ are inscribed in the same arc of $\omega$. Thus $\triangle AIF$ is isosceles with $AI = IF$. Since $I$ is the incenter of $ABC$, we have $AF = 2(s-a)$, where $s = (a+b+c)/2$ is the semiperimeter of $ABC$. Also, it is $CE = AC = b$ because in triangle $ACE$, we have
$$\begin{aligned} \angle AEC & = \angle ABC + \angle BAE \\ & = \angle ABC + \frac{\angle BAD}{2} \\ & = \angle ABC + \frac{\angle BAC - \angle ABC}{2} \\ & = 90^\circ - \frac{\angle ACE}{2} \end{aligned}$$
and so $\angle CAE = 180^\circ - \angle AEC - \angle ACE = 90^\circ - \frac{\angle ACE}{2} = \angle AEC$. Hence
$$BF = BA - AF = c - 2(s-a) = a - b = CB - CE = BE.$$
Moreover, triangle $CAX$ is similar to triangle $BFX$, because $\angle ACX = \angle FBX$ and
$$\angle XFB = \angle XAF + \angle AXF = \angle XAF + \angle CAF = \angle CAX$$
Therefore
$$\frac{CE}{BE} = \frac{AC}{BF} = \frac{XC}{XB},$$
as desired. The proof is complete.

Solution:

Let $\omega$ denote the circle through $A$ and $I$ tangent to $AC$. Let $Y$ be the second point of intersection of the circle $\omega$ with the line $AD$. Let $L$ be the intersection of $BC$ with the angle bisector of $\angle BAD$. We will prove $\angle LXC = \frac{1}{2} \angle BAC = \frac{1}{2} \angle BXC$.

We will refer to the angles of $ABC$ as $\angle A, \angle B, \angle C$. Thus $\angle BAD = \angle A - \angle B$.
On the circumcircle of $ABC$, we have $\angle AXC = \angle ABC = \angle CAD$, and since $AC$ is tangent to $\omega$, we have $\angle CAD = \angle CAY = \angle AXY$. Hence $C, X, Y$ are collinear.
Also note that $CAL$ is isosceles with $\angle CAL = \angle CLA = \frac{1}{2}(\angle BAD) + \angle ABC = \frac{1}{2}(\angle A + \angle B)$ hence $AC = CL$. Moreover, $CI$ is angle bisector to $\angle ACL$ so it's the symmetry axis for the triangle, hence $\angle ILC = \angle IAC = \frac{1}{2} \angle A$ and $\angle ALI = \angle LIA = \frac{1}{2} \angle B$. Since $AC$ is tangent to $\omega$, we have $\angle AYI = \angle IAC = \frac{1}{2} \angle A = \angle LAY + \angle ALI$. Hence $L, Y, I$ are collinear.
Since $AC$ is tangent to $\omega$, we have $\triangle CAY \sim \triangle CXA$ hence $CA^2 = CX \cdot CY$. However we proved $AC = CL$ hence $CL^2 = CX \cdot CY$. Hence $\triangle CLY \sim \triangle CXL$ and hence $\angle CXL = \angle CLY = \angle CAI = \frac{1}{2} \angle A$.

Solution:

Let $M$ be the midpoint of the $\operatorname{arc} BC$. Let $\omega$ denote the circle through $A$ and $I$ tangent to $AC$. Let $N$ be the second point of intersection of $\omega$ with $AB$ and $L$ the intersection of $BC$ with the angle bisector of $\angle BAD$. We know $\frac{DL}{LB} = \frac{AD}{AB}$ and want to prove $\frac{XB}{XC} = \frac{LB}{LC}$.

First note that $CAL$ is isosceles with $\angle CAL = \angle CLA = \frac{1}{2}(\angle BAD) + \angle ABC$ hence $AC = CL$ and $\frac{LB}{LC} = \frac{LB}{AC}$.
Now we calculate $\frac{XB}{XC}$:
Comparing angles on the circles $\omega$ and the circumcircle of $ABC$ we get $\triangle XIN \sim \triangle XMB$ and hence also $\triangle XIM \sim \triangle XNB$ (having equal angles at $X$ and proportional adjoint sides). Hence $\frac{XB}{XM} = \frac{NB}{IM}$.
Also comparing angles on the circles $\omega$ and the circumcircle of $ABC$ and using the tangent $AC$ we get $\triangle XAI \sim \triangle XCM$ and hence also $\triangle XAC \sim \triangle XIM$. Hence $\frac{XC}{XM} = \frac{AC}{IM}$.
Comparing the last two equations we get $\frac{XB}{XC} = \frac{NB}{AC}$. Comparing with $\frac{LB}{LC} = \frac{LB}{AC}$, it remains to prove $NB = LB$.

We prove $\triangle INB \equiv \triangle ILB$ as follows:
First, we note that $I$ is the circumcentre of $\triangle ALN$. Indeed, $CI$ is angle bisector in the isosceles triangle $ACL$ so it's perpendicular bisector for $AL$. As well, $\triangle IAN$ is isosceles with $\angle INA = \angle CAI = \angle IAB$ hence $I$ is also on the perpendicular bisector of $AN$.
Hence $IN = IL$ and also $\angle NIL = 2\angle NAL = \angle A - \angle B = 2\angle NIB$ (the last angle is calculated using that the exterior angle of $\triangle NIB$ is $\angle INA = \angle A/2$. Hence $\angle NIB = \angle LIB$ and $\triangle INB \equiv \triangle ILB$ by SAS.

Solution:

Let $M, N$ be the midpoints of arcs $BC, BA$ of the circumcircle $ABC$, respectively. Let $Y$ be the second intersection of $AD$ and circle $ABC$. Let $E$ be the incenter of triangle $ABY$ and note that $E$ lies on the angle bisectors of the triangle, which are the lines $YN$ (immediate), $BC$ (since $\angle CBY = \angle CAY = \angle CAD = \angle ABC$) and the angle bisector of $\angle DAB$; so the question reduces to showing that $E$ is also on $XM$, which is the angle bisector of $\angle CXB$.

We claim that the three lines $CX, ADY, IE$ are concurrent at a point $D'$. 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 $XEM$ 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 $AD'DY$ and $CIN$. Projecting the points $YNC M$ on the circle $ABC$ through the (concyclic) point $A$ to the line $CN$ yields the points $ONCI$. Projecting these points through $E$ to the line $AY$ yields $OYDD'$ (here we use the facts that $D'$ lies on $IE$ and $AY$). Projecting these points to the circle $ABC$ through $C$ yields $NYBX$ (here we use the fact that $D'$ lies on $CX$). Composing, we observe that we found a projection of the circle $ABC$ to itself sending $YNC M$ to $NYBX$. Since the projection of the circle through $E$ also sends $YNC$ to $NYB$, and three points determine a projective transformation, the projection through $E$ also sends $M$ to $X$, as claimed.

Let $B', D'$ be the intersections of $AB, AD$ with the circle $AXI$, respectively. We wish to show that this $D'$ is the concurrency point defined above, i.e. that $CD'X$ and $ID'E$ are collinear. Additionally, we will show that $I$ is the circumcenter of $AB'E$.
Consider the inversion with center $C$ and radius $CA$. The circles $AXI$ and $ABD$ are tangent to $CA$ at $A$ (the former by definition, the latter since $\angle CAD = \angle ABC$), so they are preserved under the inversion. In particular, the inversion transposes $D$ and $B$ and preserves $A$, so sends the circle $CAB$ to the line $AD$. Thus $X$, which is the second intersection of circles $ABC$ and $AXI$, is sent by the inversion to the second intersection of $AD$ and circle $AXI$, which is $D'$. In particular $CD'X$ are collinear.
In the circle $AIB'$, $AI$ is the angle bisector of $B'A$ and the tangent at $A$, so $I$ is the midpoint of the arc $AB'$, and in particular $AI = IB'$. By angle chasing, we find that $ACE$ is an isosceles triangle:
$\angle CAE = \angle CAD + \angle DAE = \angle ABC + \angle EAB = \angle ABE + \angle EAB = \angle AEB = \angle AEC$,
thus the angle bisector $CI$ is the perpendicular bisector of $AE$ and $AI = IE$. Thus $I$ is the circumcenter of $AB'E$.
We can now show that $ID'E$ are collinear by angle chasing:
$$\angle EIB' = 2\angle EAB' = 2\angle EAB = \angle DAB = \angle D'AB' = \angle D'IB'.$$

Solution:

Let $W$ be the midpoint of arc $BC$, let $D'$ be the second intersection point of $AD$ and the circle $ABC$. Let $P$ be the intersection of the angle bisector $XW$ of $\angle CXB$ with $BC$; we wish to prove that $AP$ is the angle bisector of $DAB$. Denote $\alpha = \frac{\angle CAB}{2}$, $\beta = \angle ABC$.
Let $M$ be the intersection of $AD$ and $XC$. Angle chasing finds:
$$\begin{aligned} \angle MXI & = \angle AXI - \angle AXM = \angle CAI - \angle AXC = \alpha - \beta \\ & = \angle CAI - \angle CAD = \angle DAI = \angle MAI \end{aligned}$$
And in particular $M$ is on $\omega$. By angle chasing we find
$$\angle XIA = \angle IXA + \angle XAI = \angle ICA + \angle XAI = \angle XAC = \angle XBC = \angle XBP$$
and $\angle PXB = \alpha = \angle CAI = \angle AXI$, and it follows that $\triangle XIA \sim \triangle XBP$. Let $S$ be the second intersection point of the cirumcircles of $XIA$ and $XBP$. Then by the spiral map lemma (or by the equivalent angle chasing) it follows that $ISB$ and $ASP$ are collinear.
Let $L$ be the second intersection of $\omega$ and $AB$. We want to prove that $ASP$ is the angle bisector of $DAB = \angle MAL$, i.e. that $S$ is the midpoint of the $\operatorname{arc} ML$ of $\omega$. And this follows easily from chasing angular arc lengths in $\omega$:
$$\text{AI = CAI = IL = IAL = MI = MXI = - AI - SL = ABI = 2}$$
And thus
$$\widehat{ML} = \widehat{MI} + \widehat{IL} = 2\alpha - \beta = 2\left(\widehat{AI} - \frac{\beta}{2}\right) = 2\widehat{SL}$$

Solution:

Let $E$ be the intersection of the bisector of $\angle BAD$ and $BC$, and $N$ be the middle point of arc $BC$ of the circumcircle of $ABC$. Then it suffices to show that $E$ is on line $XN$.
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 $AC^*$ are parallel. Now it suffices to show that $A, X^*, E^*, N^*$ are concyclic. Let $Y$ be the intersection of $B^*C^*$ and $AE^*$. Then, by the power of a point, we get
$$\begin{aligned} A, X^*, E^*, N^* \text{ are concyclic } &\Longleftrightarrow YX^* \cdot YN^* = YA \cdot YE^* \\ &\Longleftrightarrow YX^* \cdot YN^* = YB^* \cdot YC^* \\ &\left(A, B^*, C^*, E^* \text{ are concyclic }\right) \end{aligned}$$
Here, by the property of inversion, we have
$$\angle AI^*B^* = \angle ABI = \frac{1}{2} \angle ABC = \frac{1}{2} \angle C^*AD^*.$$

Define $Q, R$ as described in the figure, and we get by simple angle chasing
$$\angle QAI^* = \angle QI^*A, \quad \angle RAI^* = \angle B^*I^*A$$
Especially, $B^*R$ and $AI^*$ are parallel, so that we have
$$\frac{YB^*}{YN^*} = \frac{YR}{YA} = \frac{YX^*}{YC^*}$$
and the proof is completed.