Trig bash.

I started doing a solution by extending XA and YB to meet at XA and YC to meet at V and do some trig, which turns out pretty nice because P and Q are the orthocenters of WXY and VXY, but couldn't finish. Anyone want to finish for me?

You can also do it by finding 4 similar triangles and using the Pythagorean Theorem.

I solved by dropping perpendiculars from B and C to AY, but my solution might have had configuration issues.

I did similar trig bash and was a little too handwavy with a = c, but whatever. I noted configuration issues but explained that they didn't affect the result. I think I'm fine

Does anyone know if there are configuration issues that you'd have to mention?

I think B and C would coincide if you let AX and BY be 60 degree arcs. Edit - wait that doesn't really affect the outcome

Coordinate bash for the win! Ugh it was so ugly.

@fermat007 At least for my configuration there was a configuration issue depending on which side of B C was on. The angles for my trig bash ended up the same though, so I just did my initial diagram one way then attached a page in the back the other way and reproved stuff.

Hmm I didn't think of any good proof, so I just wrote some nontrivial steps that proved $\angle AXB=\angle CXY$ and $BC\parallel AY$, both of which actually seem to be important . But still I worked on this for two hours and didn't even think of trig with all those right angles >.<

Same thing as giratina, you get like the sine addition formula with gives you like sin (2a+b)/cos(2a+b) in the first diagram and that's the tangent of the angle and that's ay/ax

i think some mistake in the problem. Because, $BC\| AY.$

Why would there be a mistake in the problem? $C$ is not supposed to slide around the circle at free will. It is defined by the location of points $A$ and $B$.

you write: Let $Q$ be the intersection of segments $AY$ and $BC$ It's may not. Because, $BC\| AY$.

mathuz wrote: you write: Let $Q$ be the intersection of segments $AY$ and $BC$ It's may not. Because, $BC\| AY$. If you really want to nitpick on something like this, you should note that the definition $Q = AY \cap BC$ is part of the hypothesis. An incorrect hypothesis does not invalidate the problem statement; see the principle of explosion for an extreme case of this. In simpler words, you may happily assume $AY \nparallel BC$, because the problem tells you that they intersect at a point $Q$.

It was a typo; he meant $Q$ is the intersection of $AY$ and $XC$.

Let $\angle XYA = \alpha, \angle AYB = \beta, \angle BYC = \gamma$. $APZX$ is cyclic and $AZ, CY$ are parallel, so \[ \angle CXY = \angle YAZ \implies 90-(\alpha + \beta + \gamma) = \beta + \gamma \implies \gamma = 90 - \alpha - 2\beta \]By Ratio Lemma \[ \frac{XQ}{CQ} = \frac{XY \sin \alpha}{CY \sin (\beta + \gamma)}, CQ = CY \tan (\beta+\gamma), CY = XY \cos(\beta + \gamma) \implies XQ = XY \frac{\sin \alpha}{\cos(\beta + \gamma)} \implies \frac{CY}{XQ} = \frac{\cos (\alpha + \beta + \gamma) \cos(\beta+\gamma)}{\sin \alpha} \]\[ \frac{XP}{BP} = \frac{XY \sin \alpha}{BY \sin \beta}, BP = BY \tan \beta, BY = XY \cos(\alpha+\beta) \implies XP = XY \frac{\sin \alpha}{\cos \beta} \implies \frac{BY}{XP} = \frac{\cos (\alpha + \beta) \cos \beta}{\sin \alpha} \]Since $\frac{AY}{AX} = \frac{\cos \alpha}{\sin \alpha}$ it suffices to show \[ \cos \alpha = \cos (\alpha + \beta + \gamma) \cos(\beta+\gamma) + \cos (\alpha + \beta) \cos \beta = \cos(\alpha+\beta)\cos \beta + \cos(90-\beta) \cos (90-\alpha-\beta) \]upon plugging in our value of $\gamma$, which holds by $\cos$ subtraction identity.

Totally didn't misread $P = XA \cap BY$ then get stuck on this for two hours. [asy][asy] import olympiad; size(250); pair X = dir(180), Y = dir(0), A = dir(144), B = dir(83); pair X1 [] = intersectionpoints(A--Y, B--X); pair P = X1[0]; pair Z = foot(P, X, Y); pair X2 = extension(X, foot(X, A, Z), (50, 0), (50, 50)); pair X3 [] = intersectionpoints(X--X2, circle(origin, 1)); pair C = X3[1]; pair X4 [] = intersectionpoints(X--C, A--Y); pair Q = X4[0]; draw(arc(origin, Y, X), dashed); draw(A--Y); draw(X--B); draw(P--Z); draw(C--Y); draw(A--X); draw(X--C); draw(A--Z); draw(A--B--Y); draw(rightanglemark(A, foot(X, A, Z), X, 1)); draw(X--Y); dot("$A$", A, NW); dot("$B$", B, NE); dot("$X$", X, SW); dot("$Y$", Y, SE); dot("$Z$", Z, S); dot("$Q$", Q, N); dot("$P$", P, N); dot("$C$", C, NE); [/asy][/asy] WLOG let the radius of the circle be $\frac 12$, and let $\angle AXY = \alpha$ and $\angle XYB = \beta$. Then we have \begin{align*} BY &= \cos \beta \\ \angle AXB = \alpha+\beta-90^{\circ} \implies XP &= \frac{AX}{\sin(\alpha+\beta)} = \frac{\cos \alpha}{\sin(\alpha+\beta)}. \end{align*}Next, verify that since $APZX$ is cyclic, $\angle CYA = 90^{\circ} - \beta$, so $\angle CYX = 180^{\circ} - \alpha - \beta$. This gives us $$CY = -\cos(\alpha+\beta),$$and since $\angle AXQ = \angle QAZ = 90^{\circ} - \beta$, $$XQ = \frac{AX}{\sin \beta} = \frac{\cos \alpha}{\sin \beta}.$$It is also straightforward to find that $AY=\sin \alpha, AX = \cos \alpha$. Then the desired becomes \begin{align*} \iff \frac{\frac{\cos \beta}{\cos \alpha}}{\sin(\alpha+\beta)} - \frac{\cos(\alpha+\beta)}{\frac{\cos \alpha}{\sin \beta}} &= \frac{\sin \alpha}{\cos \alpha} \\ \iff \cos \beta \sin (\alpha+\beta)-\sin \beta \cos(\alpha+\beta) &= \sin \alpha, \end{align*}which is obvious by sine subtraction. We are done. $\square$

Let $XA$ and $YB$ intersect at $K$. By Pascal's on $XXAYYB$, $K$, $P$, and $Z$ are collinear. In particular, $P$ is the orthocenter of $\triangle KXY$. Moreover, since $AZ$ is antiparallel to $KY$, $XQ$ is isogonal to $XP$. We may now compute, in terms of the side lengths and angles of triangle $KXY$, $BY = k\cos{Y}$ $XP = \frac{XZ}{\cos{\angle PXZ}} = \frac{y\cos{X}}{\sin{Y}} = \frac{k\cos{X}}{\sin{K}}$ $CY = k\cos{\angle CYX} = k\cos{K}$ $XQ = \frac{XA}{\cos{\angle AXQ}} = \frac{k\cos{X}}{\sin{Y}}$ and $\frac{BY}{XP} +\frac{CY}{XQ} = \frac{\cos{Y}\sin{K}}{\cos{X}} + \frac{\cos{K}\sin{Y}}{\cos{X}} = \frac{\sin{X}}{\cos{X}} = \frac{AY}{AX}$ as desired.

It is obvious that $\triangle XAY$, $\triangle XBY$ and $\triangle XZY$ are right-angled at $A,B,C$. We now define $\measuredangle YXB\coloneq \measuredangle 1$, $\measuredangle AYX\coloneq \measuredangle 2$, $\measuredangle BXA\coloneq \measuredangle 3$, $\measuredangle APX\coloneq \measuredangle 4$, $\measuredangle AQX\coloneq \measuredangle 5$, $\measuredangle YXC\coloneq \measuredangle 6$. We now find that, in $\triangle XAP$, $$XP=\frac{AX}{\sin(\measuredangle 4)}$$by the law of sines. Similarly in $\triangle XAQ$: $$XQ=\frac{AX}{\sin(\measuredangle 5)}$$If we plug that in into the desired equation and simplify, it is now enough to prove that $$\frac{BY\sin(\measuredangle 4)}{AX}+\frac{CY\sin(\measuredangle 5)}{AX}=\frac{AY}{AX} \Longleftrightarrow BY\sin(\measuredangle 4)+CY\sin(\measuredangle 5)=AY\quad (1)$$We now express $$BY=XY\sin(\measuredangle 1)\quad CY=XY\sin(\measuredangle 5)$$by applying the law of sines in $\triangle XBY$ and $\triangle XCY$. Furthermore, we have $AY=XY\sin(\measuredangle 1+\measuredangle 3)$ in $\triangle XAY$. Pluggin that into (1) and dividing by $XY$ gives $$\sin(\measuredangle 1)\sin(\measuredangle 4)+\sin(\measuredangle 6)\sin(\measuredangle 5)=\sin(\measuredangle 1+\measuredangle 3)=\cos(\measuredangle 2)$$upon noticing that $\measuredangle 2=90^\circ-(\measuredangle 1+\measuredangle 3)$ in triangle $\triangle XAY$ by the interior angle sum. We now simplify the sine products using $$\sin(x)\sin(y)=\frac{1}{2}(\cos(x-y)-\cos(x+y))$$We get: $$\sin(\measuredangle 1)\sin(\measuredangle 4)=\frac{1}{2}(\cos(\measuredangle 1-\measuredangle 4)-\cos(\measuredangle 1+\measuredangle 4))=\frac{1}{2}(\cos(\measuredangle 2)-\cos(2\measuredangle 1+\measuredangle 2))$$because $\measuredangle 1+\measuredangle 2=\measuredangle 4$ using the exterior angle sum formula in $\triangle XPY$. Similarly, we get $$\sin(\measuredangle 5)\sin(\measuredangle 6)=\frac{1}{2}(\cos(\measuredangle 2)-\cos(2\measuredangle 6+\measuredangle 2))$$Summing those, subtracting $\cos(\measuredangle 2)$ and diving by $\frac{-1}{2}$ gives: $$\cos(2\measuredangle 1+\measuredangle 2)+\cos(2\measuredangle 6+\measuredangle 2)=0$$Since $-\cos(x)=\cos(180^\circ-x)$, it is enough to prove that $$2\measuredangle 1+\measuredangle 2=180^\circ-(2\measuredangle 6+\measuredangle 2)\Longleftrightarrow 2\measuredangle 1+2\measuredangle 6+2\measuredangle 1=180^\circ$$However, since $PZ\perp XY$, $APZX$ is cyclic, so $\measuredangle PXA=\measuredangle PZA=\measuredangle 3$. However, $AZ\perp XQ$. It is well-known and easy to prove by angle chasing that, if we let $F=AZ\cap XQ$ and $G=ZP\cap XQ$, we then have $\measuredangle 3=\measuredangle PZA=\measuredangle GZF\measuredangle GZF=\measuredangle ZXQ=\measuredangle 6$. To conclude the proof, let $\ell$ be the tangent to $\omega$ at $X$. By the tangent-chord theorem, $\measuredangle 2=\measuredangle AYX=\measuredangle (AX,\ell)$. We now see that $$\measuredangle ZXP+\measuredangle PXA+\measuredangle (AX,\ell)=\measuredangle 1+\measuredangle 3+\measuredangle 2=\measuredangle 1+\measuredangle 6+\measuredangle 2=90^\circ$$. Multiplying by two and working backwards gives the desired result.

It's clear that $AX, BY, PZ$ meet at the orthocenter of $PXY$, which we denote by $H$. Since $AHYZ$ is cyclic, $AZ$ and $YH$ are anti-parallel wrt $XH$ and $XY$. Now, because Thales' implies $XB \perp YH$ and we're given $XC \perp AZ$, it follows that $XB$ and $XC$ are isogonal in $\angle HXY$ or $\angle AXY$. Thus, we have $AY \parallel BC$, so $AYBC$ is an isosceles trapezoid. Let $E, F$ denote the projections of $A, Y$ onto $BC$ respectively. Observe $$\angle YAX = 90^{\circ} = \angle BEA = 180^{\circ} - \angle YAE$$so $E$ lies on $AX$. Isogonality and angle chasing gives $XAQ \overset{+}{\sim} XBY$ and $XAP \overset{+}{\sim} XCY$, so $$\frac{XQ}{XY} = \frac{AQ}{BY} \implies XQ \cdot BY = XY \cdot AQ,$$$$\frac{XP}{XY} = \frac{AP}{CY} \implies XP \cdot CY = XY \cdot AP,$$and $$\frac{XA}{XC} = \frac{XP}{XY} \implies XA \cdot XY = XC \cdot XP.$$Hence, $$\frac{BY}{XP} + \frac{CY}{XQ} = \frac{XQ \cdot BY + XP \cdot CY}{XP \cdot XQ} = \frac{XY(AQ + AP)}{XP \cdot XQ }= \frac{(XA \cdot XY)(AQ + AP)}{XA \cdot XP \cdot XQ}$$$$= \frac{(XC \cdot XP)(AQ + AP)}{XA \cdot XP \cdot XQ} = \frac{XC}{XQ} \cdot (AQ + AP) \cdot \frac{1}{XA}$$$$= \frac{XE}{XA} \cdot (AQ + AP) \cdot \frac{1}{AX} = (EC + EB) \cdot \frac{1}{AX} = \frac{EC + FC}{AX}$$$$= \frac{EF}{AX} = \frac{AY}{AX}$$as desired. $\blacksquare$ Remarks: Initially, the condition $XC \perp AZ$ seems quite strange. Once we realize the diagram is an orthocenter configuration, however, it becomes much easier to recognize the isogonal lines, allowing us to find similar triangles and substitute for certain length expressions.

Trig is amazing. Denote $\angle BXY = a$ and $\angle AYX = b$. WLOG, let $XY=1$. Then $BY = \sin a, AX = \sin b, BX = \cos a, AY = \cos b$. Ptolemy's on $XABY$ gives $AB = \cos a \cos b - \sin a \sin b = \cos (a+b)$. Using similar triangles within cyclic quadrilateral $XABY$, we find that $PX/ PB = \frac{\sin b}{\sin a \cos (a+b)}$. Since $PX+PB = BX = \cos a$, we can solve for $XP$ to be $\frac{\sin b \cos a}{\sin a \cos (a+b) + \sin b}$. Then $$BY/XP = \frac{\sin a}{\frac{\sin b \cos a}{\sin a \cos (a+b) + \sin b}} = \frac{\sin^2 a \cos (a+b) + \sin a \sin b}{\sin b \cos a}.$$To find $CY/XQ$, we can do the same thing, except this time, we have $\angle CXY$ instead of $a$. We can see that $XAPZ$ is cyclic since $\angle XAP = \angle XZP = 90$, so $\angle CXY = 90 - \angle AZX = 90 - \angle APX = 90 - (a+b).$ Now we can replace $90-(a+b)$ for $a$ in the previous expression and repeatedly using the fact that $\sin \theta = \cos (90 - \theta)$, we get that $$CY / XQ = \frac{\cos^2 (a+b) \sin a + \cos(a+b) \sin b}{ \sin b \sin(a+b)}.$$Adding these, we get that \begin{align*} BY/XP + CY/XQ &= \frac{\sin^2 a \cos (a+b) + \sin a \sin b}{\sin b \cos a} + \frac{\cos^2 (a+b) \sin a + \cos(a+b) \sin b}{ \sin b \sin(a+b)} \\ &= \frac{\sin a \sin(a+b) (\sin a \cos(a+b) + \sin b) + \cos(a+b) \cos a (\cos (a+b) \sin a + \sin b) }{\sin (a+b) \cos a \sin b}\\ &= \frac{(\sin a \sin (a+b) + \cos a \cos(a+b)) (\sin a \cos(a+b) + \sin b)}{\sin(a+b)\cos a \sin b} \\ &= \frac{\cos(a-(a+b)) (\sin (a+b) \cos a)}{ \sin(a+b) \cos a \sin b } ^{*} \\ &= \frac{\cos b}{\sin b}\\ &= \frac{AY}{AX}. \end{align*} The simplifications on $^{*}$: $\sin a \sin(a+b) + \cos a \cos (a+b) = \cos(a-(a+b))$ follows from the cosine subtraction formula. $\sin a \cos(a+b) + \sin b = \frac 12 (\sin (2a+b) - \sin b) + \sin b = \frac 12 (\sin(2a+b) + \sin b) = \sin(a+b) \cos (a)$ by the product - sum and sum-product rules.

you haven't seen real trig until you do 2019 tstst 9

Boring trig bash. Let $W=AX\cap BY$, $\alpha = \angle XWZ$, and $\beta = \angle YWZ$. Then $P$ is the orthocenter of triangle $WXY$ and the feet of the altitudes are $A$, $B$, and $Z$. WLOG let $WZ=1$. Perform standard angle chasing and length bashing with the orthocenter configuration to obtain \begin{align*} BY &= (\tan\alpha + \tan\beta)\sin\beta \\ XP &= \frac{\tan\alpha}{\cos\beta} \\ CY &= (\tan\alpha + \tan\beta)\cos(\alpha+\beta) \\ XQ &= \frac{(\tan\alpha + \tan\beta)\sin\alpha}{\cos\beta} \\ \frac{AY}{AX} &= \frac{1}{\tan\alpha} \end{align*}Now it suffices to show that \begin{align*} \frac{(\tan\alpha + \tan\beta)\sin\beta}{\tan\alpha}\cdot\cos\beta + \frac{\cos(\alpha+\beta)\cos\beta}{\sin\alpha} &= \frac{1}{\tan\alpha} \\ (\tan\alpha+\tan\beta)\sin\beta\cos\beta + \frac{\cos(\alpha+\beta)\cos\beta}{\cos\alpha} &= 1 \\ \sin\alpha\sin\beta\cos\beta + \sin^2\beta\cos\alpha + \cos(\alpha+\beta)\cos\beta &= \cos\alpha \\ \sin\alpha\sin\beta\cos\beta + \sin^2\beta\cos\alpha + (\cos\alpha\cos\beta - \sin\alpha\sin\beta)\cos\beta &= \cos\alpha \\ (\sin^2\beta + \cos^2\beta)\cos\alpha &= \cos\alpha \end{align*}which is true.

Let $\angle AYX=a$ and $\angle BXY=b$. Note that since $\angle XAY=90$ (semicircle), we have that $AXZP$ is cyclic (since $\angle PZX=90$), which means that \[\angle ZAP=\angle ZXP=b,\]which in turn gives that \[\angle XAZ=90-\angle ZAY=90-b,\]and \[\angle AXQ=90-\angle XAZ=b,\]which is all we need to do our trigbash. By Law of Sines, we have that \[BY=XY*\frac{\sin b}{\sin 90},\]on $\triangle BXY$ and \[XP=XY*\frac{\sin a}{\sin (a+b)},\]on $\triangle XPY$. These two give us that \[\frac{BY}{XP}=\frac{\sin b \sin (a+b)}{\sin a}.\] Similarly, by Law of Sines, we also have that \[CY=XY*\frac{\sin (90-(a+b))}{\sin 90}=XY*\frac{\cos (a+b)}{\sin 90},\]on $\triangle CXY$ and \[XQ=XY*\frac{\sin a}{\sin (90+b)}=XY*\frac{\sin a}{\sin (90-b)}=XY*\frac{\sin a}{\cos b},\]on $\triangle QXY$. These two equations give us that \[\frac{CY}{XQ}=\frac{\cos b \cos (a+b)}{\sin a}.\] Adding gives that \[\frac{BY}{XP}+\frac{CY}{XQ}=\frac{\sin b \sin (a+b)}{\sin a}+\frac{\cos b \cos (a+b)}{\sin a}=\frac{cos((a+b)-b)}{\sin a}=\frac{\cos a}{\sin a},\]which is indeed equal to $\frac{AY}{AX}$, since $\triangle AYX$ is a right triangle with $\angle XAY=90$ and $\angle XYA=a$, finishing the problem.

Let angles $K$, $X$, and $Y$ represent the angles of $\triangle KXY$, where $K = AX \cap BY \cap PZ$. Our problem is essentially 2 orthic configurations, and we can find $\angle CXY = 90 - \angle K$. Then the LHS can be expressed as \begin{align*} \frac{BY}{XP} + \frac{CY}{XQ} &= \frac{BY}{XY \cdot \frac{\cos X}{\sin K}} + \frac{CY}{XY \cdot \frac{\cos X}{\sin Y}} \\ &= \frac{BY}{\frac{BY}{\cos Y} \cdot \frac{\cos X}{\sin K}} + \frac{CY}{\frac{CY}{\cos K} \cdot \frac{\cos X}{\sin Y}} \\ &= \frac{\cos Y \sin K + \cos K \sin Y}{\cos X} \\ &= \frac{\sin X}{\cos X} = \frac{AY}{AX}. \quad \blacksquare \end{align*}

Let $\angle AYX = \theta$ and $\angle BXY = \alpha,$ so $\angle APX = \alpha + \theta$ by inscribed arc properties. We now contend that $ABCY$ is an isosceles trapezoid and therefore $AB = CY$ and $AC = BY.$ First, $APXZ$ is cyclic, since $\angle PAX = \angle PZX = 90^\circ,$ and similarly $BPZY$ is cyclic. Thus $$\angle AXC = 90^\circ - \angle XAZ = 90^\circ - \angle XPZ = 90^\circ + \angle BPZ = 90^\circ - \angle BYZ = \angle BXY,$$implying the claim. We now proceed to the problem. First, we have $XP = \frac{XA}{\sin \angle XPA} = \frac{XA}{\sin(\theta+\alpha)},$ so $\frac{BY}{XP} = \frac{BY}{XA} \cdot \sin(\theta+\alpha).$ By the Law of Sines, $\frac{BY}{XA} = \frac{\sin \alpha}{\sin \theta},$ so $$\boxed{\frac{BY}{XP} = \frac{\sin \alpha \sin(\theta + \alpha)}{\sin \theta}}.$$ Next, we have $\angle AXC = \angle BXY = \alpha$ by our claim, thus $XQ = \frac{XA}{\cos \angle AXQ} = \frac{AX}{\cos \alpha}.$ Hence $\frac{CY}{XQ} = \frac{CY}{XA} \cdot \cos \alpha.$ By the Law of Sines, $\frac{CY}{XA} = \frac{\sin \angle CXY}{\sin \angle XYA} = \frac{\sin(90^\circ - \alpha - \theta)}{\sin \theta} = \frac{\cos(\alpha + \theta)}{\sin \theta}$ since $\angle CXY = \angle XAB = 90^\circ - \alpha - \theta$, once again by our claim. Therefore, $$\boxed{\frac{CY}{XQ} = \frac{\cos \alpha \cos(\theta + \alpha)}{\sin \theta}}.$$ The problem is straightforward from here. Adding the two boxed expressions up gives $$\frac{BY}{XP} + \frac{CY}{XQ} = \frac{\sin \alpha \sin (\theta + \alpha) + \cos \alpha \cos (\theta+\alpha)}{\sin \theta} = \frac{\cos \theta}{\sin \theta} = \cot \angle AYX = \frac{AY}{AX}$$by the cosine angle subtraction formula, hence we are done.

Claim. $\angle AXQ = \angle PXY$ Proof. First, note that $\angle XAP = \angle XAY = 90^{\circ}$, so $XAPZ$ is cyclic. It follows that \begin{align*} \angle AXQ &= 90^{\circ} - \angle AQX \\ &= \angle ZAP \\ &= \angle PXZ \\ &= \angle PXY \end{align*}as desired. $\blacksquare$ Now, the given condition is equivalent to \begin{align*} \dfrac{AX \cdot BY}{XP} + \dfrac{AX \cdot CY}{XQ} &= AY \\ \iff BY \sin APX + CY \sin AQX &= AY \\ \iff BY \sin XPY + CY \sin XQY &= AY \end{align*}By the Law of Sines in $\Delta XPY$, we have $$\sin XPY = \dfrac{XY \sin PXY}{PY}.$$Applying the Law of Sines in $\Delta XQY$ yields $$\sin XQY = \dfrac{XY \sin QXY}{QY}$$Thus, we have \begin{align*} BY \sin XPY + CY \sin XQY &= XY \left(\dfrac{BY \sin PXY}{PY} + \dfrac{CY \sin QXY}{QY}\right) \\ &= XY \left(\dfrac{BY \sin QXA}{PY} + \dfrac{CY \sin QXY}{QY}\right) \\ &= XY (\sin QXA \cdot \cos PYB + \sin QXY \cdot \cos CYQ) \\ &= XY (\sin QXA \cdot \cos PXA + \sin QXY \cdot \cos QXA) \\ &= XY (\sin QXA \cdot \cos QXY + \sin QXY \cdot \cos QXA) \\ &= XY \cdot \sin(QXA + QXY) \\ &= XY \cdot \sin AXY \\ &= XY \cdot \dfrac{AY}{XY} \\ &= AY \end{align*}and we are done. $\blacksquare$

$ $