Can someone check if the problem statement is correct? This didn't seem to be true when I drew it with geogebra...

Here's the original statement.

@above, it should be mentioned that $ABCD$ is cyclic, which you didn't write.

XbenX wrote: @above, it should be mentioned that $ABCD$ is cyclic, which you didn't write. I think the quadrilateral $ABCD$ should be circumscribed, not cyclic.

yeah, you're right.

I got a really weird solution. First we prove (not so weirdly) that the incenter $I$ of $ABCD$ lies in the incircle of $DXY$. We do that by showing that $\angle BXD = \frac12(\angle BAD - \angle DCA)$ (which is https://artofproblemsolving.com/community/c6h2132843p15602364). Stewart's theorem on $ABC$ (or the median formula) shows that $BX^2 = AB\cdot BC$: in fact, $BX^2=\frac14(2AB^2+2BC^2-AC^2) = \frac14(2AB^2+2BC^2-2(BC-BA)^2)=AB\cdot BC.$ Relations like $BX^2 = BA\cdot BC$ reminds us of power of a point. Indeed, consider point $A'$ on side $BC$ such that $BA'=BA$. Then $BA'\cdot BC=BX^2$ shows that $BX$ is tangent to the circumcircle of $CXA'$. Then $\angle A'XB = \angle BCA$. We also have $CA' = BC-BA=AC/\sqrt2$, so $CX\cdot CA = AC^2/2=CA'^2$, and $CA'$ is tangent to the circumcircle of $A'XA$, and $\angle A'XA = \angle BA'A = 90^\circ - \frac12\angle ABC$. Thus $\angle BXA = \angle A'XA - \angle A'XB = 90^\circ - \frac12\angle ABC - \angle BCA = \frac12(\angle BAC - \angle BAC)$. Noting that $CD-AD=BC-BA$, we can see that triangle $ADC$ also has $\sqrt2(CD-AD)=AC$, so $\angle DXA = \frac12(\angle DAC - \angle DCA)$. Summing these two inequalities yield the partial result. Then $\angle DXY = \frac12(\angle BAD-\angle BCD)$, and $\angle DIY = 180^\circ-\angle BID=180^\circ-(360^\circ-\frac12\angle ABC-\frac12\angle CDA-\angle BAD)=180^\circ-(\frac12\angle ABC + \frac12\angle CDA+\angle BCD)=\frac12(\angle BAD-\angle BCD)=\angle DXY$, and $DIXY$ is cyclic. Now we can remove point $Y$ from the problem, and it suffices to prove that $\angle BDI = \angle DXI$. Now the weird part starts. Circumscribed quadrilaterals induce quite naturally hyperbolas because $AB+CD=AD+BC\iff BC-BA=DC-DA$ (we have used this relation before). Then $B$ and $D$ are on the same branch of a hyperbola with foci $A$ and $C$. Its eccentricity is $\frac{AC}{BC-BA}=\sqrt2$, so it's an equilateral hyperbola. So we can assume that its equation is $y=1/x$ and coordinate bash from now on. Notice that we don't need points $A$ and $C$ anymore. We have $X=(0,0)$, $B=\left(b,\frac1b\right)$, and $D=\left(d,\frac1d\right)$. The bisector of $\angle ABC$ is the line that passes through $B$ and is tangent to the hyperbola (it's a well known property), so it has equation $y-\frac1b = -\frac1{b^2}(x-b)$, which reduces to $y=-\frac1{b^2}x+\frac2b$. The bisector of $\angle CDA$ is $y=-\frac1{d^2}x+\frac2d$. Point $I$ is the intersection of these bisectors, which after some not so extensive algebra, is $\left(\frac{2bd}{b+d},\frac2{b+d}\right)$. Now we are ready to finish the problem: denote by $m_\ell$ the slope of line $\ell$. Then $\bullet$ $m_{DI} = -\frac1{d^2}$ because $DI$ is tangent to the hyperbola; $\bullet$ $m_{BD} = \frac{\frac1b-\frac1d}{b-d}=-\frac1{bd}$; $\bullet$ $m_{XD} = \frac{1/d-0}{d-0}=\frac1{d^2}=-m_{DI}$; $\bullet$ $m_{XI} = \frac{2/(b+d)-0}{2bd/(b+d)-0}=\frac1{bd}=-m_{BD}$. and then we are done because $\tan\angle DXI = \frac{m_{XI}-m_{XD}}{1+m_{XI}m_{BD}} = \frac{-m_{BD}+m_{DI}}{1+m_{BD}m_{DI}}=\tan\angle IDB$. I'm curious to see any purely synthetic solution. I like this analytic approach, but I want to know how this problem was invented.

I think you can finish the problem easier ( without using the weird part) , since I am lazy I let the readers prove the following lemma. We need to prove that : $\angle{IXD}=\angle{IDB}$ There is a well known lemma that states : in a circumscribed quadrilateral the midpoints of the diagonals are collinear with the center of the inscribed circle. This gives us that $XI$ passes through the midpoint of $DB$ so in other word $XI$ is the median of $\triangle{BXD}$. We know that $\angle{BID}=180^{\circ}-\angle{DIY}=180^{\circ}-\angle{BXD} $ Statement: For a point like $S$ on the median of $AM$ in a triangle $\triangle{ABC}$ which the angle $\angle{BSC}=180^{\circ}-\angle{A}$ we have that : $MB^2 =MX \cdot MA$ $$\implies \angle{SBM}=\angle{MAB} \quad \text{and} \quad \angle{SCM}=\angle{MAC}$$Again cause I'm lazy and it is not hard to prove I will leave the proof to the readers. In this problem we have that $I$ is a point like $S$ on the median of triangle $\triangle{XBD}$ $$\implies \angle {IXD}=\angle{IDB}$$$Q.E.D$