Let the incircle $(I)$ touch $BC$ at $Z.$ $XY$ and $BC$ are the polars of $A,Z$ with respect to $(I)$ $\Longrightarrow$ $D \equiv XY \cap BC$ is the pole of $AZ$ with respect to $(I)$ $\Longrightarrow$ $ID \perp AZ.$ Since $DA$ is tangent to the circumcircle $(O)$ and the cross ratio $(B,C,D,Z)$ is harmonic, then we deduce that $AZ$ is also the polar of $D$ with respect to $(O)$ $\Longrightarrow$ $OD \perp AZ.$ Therefore, $O,I,D$ lie on a perpendicular to $AZ.$

really beautiful luis, I couldn't believe it the moment I was reading because I used a 3page complete brute force for solving it

An interesting metrical remark. IRAN, 2011. In $\triangle ABC$ let $X$ and $Y$ be the tangent points of incircle $w=C(I,r)$ with $AB$ and $AC$ respectively. The tangent at $A$ to the circumcircle $C(O,R)$ of $\triangle ABC$ intersects $BC$ at $D$ . Prove that $D\in XY\iff D\in OI\iff IL\parallel BC\iff\ IO\perp AZ$ $ \iff$ $\ a(b+c)=b^2+c^2$ $\iff (s-a)^2=(s-b)(s-c)$ where $2s=a+b+c$ , $Z\in BC\cap w$ and $L$ is the Lemoine's point (symmedian center) . Proof. Suppose w.l.o.g. $b\ne c$ . $\blacktriangleright\ \boxed{D\in XY}\iff$ $\frac {DB}{DC}=\frac{ZB}{ZC}\iff$ $\frac {c^2}{b^2}=\frac {s-b}{s-c}\iff$ $\frac {b^2-c^2}{b^2+c^2}=\frac {b-c}a\iff$ $\frac {b+c}{b^2+c^2}=\frac 1a\iff$ ${\boxed{a(b+c)=b^2+c^2}}$ . $\blacktriangleright\ \boxed{IL\parallel BC}\iff$ $[BLC]=[BIC]\iff$ $\frac {a^2}{a^2+b^2+c^2}=\frac {a}{2s}\iff$ $a^2(a+b+c)=a\left(a^2+b^2+c^2\right)\iff$ $\boxed{a(b+c)=b^2+c^2}$ . $\blacktriangleright\ \boxed{(s-a)^2=(s-b)(s-c)}\iff$ $a^2-2a(b+c)+(b+c)^2=a^2-(b-c)^2\iff$ $\boxed{a(b+c)=b^2+c^2}$ . $\blacktriangleright\ \boxed{IO\perp AZ}\iff$ $OA^2-OZ^2=IA^2-IZ^2\iff$ $ZB\cdot ZC=IA^2-IX^2\iff$ $\boxed{(s-b)(s-c)=(s-a)^2}$ . $\blacktriangleright\ \boxed{D\in XY}\iff$ $Z$ is the conjugate of $D$ w.r.t. $\{B,C\}$ $\iff$ $AZ$ is polar line of $D \iff$ $DO\perp AZ$ $\iff \boxed{D\in OI}$ .

oh..nice problem...and nice solution by me lemma 1:suppose that $AZ$intersect incircle at $P$draw tangent to incircle at $P$proof that $D$lie on this tangent..!proof:its not hard and im sure that see it in mathlink if U need say me i message it lemma 2$\frac{AB^2}{AC^2}=\frac{BZ}{ZC}$that $Z$ is the point that incircle is tangent to $BC$ proof: $\frac{BZ}{ZC}=\frac{BD}{DC}=\frac{DA^2}{DC^2}=\frac{{\sin{C}}^2}{{\sin{DAC}}^2}=\frac{{\sin{C}}^2}{{\sin{B}}^2}=\frac{AB^2}{AC^2}$ proof or problem: we proof $IDC=ODC$ first we have:($H$is the perpendicular of $A$on $BC$and $M$ is midpoint of $BC$ $IDC=IPZ=IZP=ZAH$ for second we have: $ODC=OAM$ we know that: $MAC=ZAB$(because that we proof by lemma 2 $AZ$is symmedian!)and$OAC=HAB\rightarrow OAM=ZAH \rightarrow IDC=ODC$

If $D,X,Y$ are collinear,then simple angle chasing yeilds that $\angle{ADC}$ is bisected by $DXY$.So we have $\frac{DB}{DA}=\frac{BX}{AX}=\frac{s-b}{s-a}$ and $\frac{DC}{DA}=\frac{CY}{AY}=\frac{s-c}{s-a}$.Multiplying these two relations and noting that $DA^2=DB \times DC$ we get $(s-a)^2=(s-b)(s-c) \Rightarrow {\triangle}^2=(s-a)^3$.Now let $IZ \perp DC$ and $OK \perp DC$ with $Z,K$ on $BC$.Then sine rule in $\triangle{DBA}$ gives $\frac{DB}{c}=\frac{sinC}{sin(B-C)} \Rightarrow DB=\frac{csinC}{sin(B-C)}=\frac{\frac{c^2}{2R}}{\frac{b}{2R}\frac{a^2+b^2-c^2}{2ab}-\frac{c}{2R}\frac{a^2+c^2-b^2}{2ac}}=\frac{c^2a}{b^2-c^2}$. $\frac{IZ}{OK}=\frac{r}{RcosA}=\frac{\frac{\triangle}{s}}{\frac{abc}{4\triangle}\frac{c^2+b^2-a^2}{2bc}}=\frac{2(b+c-a)^3}{a(c^2+b^2-a^2)(a+b+c)}$.(Here we have used the fact that ${{\triangle}^2=(s-a)^3}$). Now $\frac{DZ}{DK}=\frac{IZ}{OK} \Leftrightarrow \frac{DB+(s-b)}{DB+\frac{a}{2}}=\frac{2(b+c-a)^3}{a(c^2+b^2-a^2)(a+b+c)} \Leftrightarrow DB=\frac{c^2a}{b^2-c^2}$ which is true.So $\frac{DZ}{DK}=\frac{IZ}{OK}$ and $D,I,O$ are collinear. I think the statement can be 'if and only if' because both the conditions yeild $(s-a)^2=(s-b)(s-c)$.Anyways credit goes to Luis for his projective approach.