Let $W,X,Y,Z$ be the tangency points. Invert with respect to the incircle. We obtain the following equivalent problem: Quote: Let $I$ be the circumcenter of cyclic quadrilateral $WXYZ$, and let $A,B,C,D$ be the midpoints of $XY,YZ,ZW,WX$ respectively. Let $O_1,O_2$ be the reflections of $I$ in lines $AD$ and $CD$, and let $O_3$ be the reflection of $I$ in line $O_1O_2$. Prove that $B,I,O_3$ are collinear. Let $E,F$ be the feet of $I$ onto $AD$ and $CD$. Clearly it is equivalent to prove that line $EF$ is parallel to line $XY$ (since $IB \perp XY$ and $IO_3 \perp O_1O_2 \parallel EF$). [asy][asy] size(8cm); pair X = dir(140), Y = dir(70), W = dir(210), Z = dir(-30), A = (X+W)/2, D = (W+Z)/2, C = (Z+Y)/2, I = origin, E = foot(I, A, D), F = foot(I, C, D); draw(W--X--Y--Z--cycle, heavygreen); draw(unitcircle, deepgreen); draw(circumcircle(I, E, F), deepgreen); draw(W--Y^^X--Z, olive); draw(A--D--C, olive); draw(E--F, heavygreen); draw(E--I--F, lightolive); string[] names = {"$A$", "$C$", "$D$", "$E$", "$F$", "$I$", "$W$", "$X$", "$Y$", "$Z$"}; pair[] points = {A, C, D, E, F, I, W, X, Y, Z}; pair[] ll = {A, C, D, E, F, dir(130), W, X, Y, Z}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] This is just angle chasing: \begin{align*} \angle (EF, XY) &= \angle (EF, IE) + \angle (IE, AD) + \angle (AD, XY) \\ &= \angle FDI + 90^\circ+ \angle ZXY \\ &= \angle (FD, WZ) + \angle (WZ, DI) + 90^\circ+ \angle ZXY \\ &= \angle YWZ + \angle ZXY \\ &= 0. \end{align*}

Here is a short solution without inversion: Equivalently, we need to prove that the tangent line to $(O_1IO_2)$ at $I$ is perpendicular to $BI$, in other words $\angle BIO_2=90^\circ+\angle IO_1O_2$. But it is an easy angle chase to see that both sides are equal to $90^\circ+\frac{\alpha}{2}$.

Here's the short sketch of my solution: 1. notice $IO_1DO_2$ is a deltoid 2. by angle chasing, show that quadrilateral $IO_1DO_2$ is also cyclic. From this statement we get that circumcenter of $\triangle O_1IO_2$ is the midpoint of $O_1O_2$. Call this point M. So for now, if we prove that points $B, I, M$ are collinear, we're done. 3. Prove the statement above by angle chasing: Call the intersection of $ID$ and $O_1O_2$ point $P$. Use well known lemma to get $ \angle O_1IM = \angle O_2IP$ By some calculations, get $ \angle AIO_1 = \angle ABI$ and $\angle BAI = \angle IAD = \angle IO_1O_2 = \angle O_1IM $ $ \angle BIA + \angle AIO_1 + \angle O_1IM = 180^{\circ} - \angle ABI - \angle BAI + \angle ABI + \angle BAI = 180^{\circ} $ So we're done.