A medial line parallel to $AC$ of $\triangle ABC$ meets its cicumcircle at $X$ and $Y$. Let $I$ be the incentre of $\triangle ABC$ and $D$ be the mid-point of $\overset{\frown}{AC}$ not containing $B$. $L$ lies on $DI$ in such a way that $DL = BI/2$. Prove that $\angle IXL = \angle IYL$.
I was able to show that $D, I, B$ are collinear. I also proved the statement for $AB = BC$, as $BI$ becomes the perpendicular bisector of $XY$.
However, I am lost in the general case, as I am unable to find any property of $L$.
Image for reference:
Best Answer
I propose another solution. Let's first state two facts that we will use in the proof.
Characterization of inscribed quadrilaterals. If two lines, one containing segment $A_1A_2$ and the other containing segment $B_1B_2$, intersect at $K$, then the four points $A_1, A_2, B_1, B_2$ are concyclic if and only if $$ KA_1\cdot KA_2=KB_1\cdot KB_2. $$
Suppose $\triangle ABC$ has an incircle with center $I$ and an excircle tangent to $AC$ with center $J_B$. Then $$ BI\cdot BJ_B=AB\cdot BC.\tag1 $$ Here is a short proof of this statement (see the figure on the left). Since $\angle ABI=\angle CBJ_B$ and $\angle AIB=\angle J_BCB$, it follows that the triangles $AIB$ and $J_BCB$ are similar and $(1)$ is proven.
Let the line $BJ_B$ intersect the circumcircle of triangle $ABC$ at a point $D$. With the same notations as in the previous item, the equality is true: $$ \frac{BJ_B}{2}=BD-\frac{BI}{2}. $$ Check this for yourself.
Now let's prove the statement from the question (see the figure on the right).
Reflecting $X$ about the bisector of angle $B$, we obtain a point $X'\in BY$. We have $\angle IXL=\angle IX'L$. If we will prove that $IX'YL$ is inscribed quadrilateral, then $\angle IX'L=\angle IYL$.
It follows from item 2 that the equality $(1)$ holds. Let the lines $BX$ and $CA$ intersect at the point $Z$. Since $\triangle ABZ$ and $\triangle YBC$ are similar, it follows that $AB\cdot BC=BZ\cdot BY$. Hence according item 3 we get \begin{align*} 2BX\cdot BY=BI\cdot BJ_B &\\ &\Longrightarrow 2BX'\cdot BY=BI\cdot BJ_B\\ &\Longrightarrow 2BX'\cdot BY=BI\cdot2BL\\ &\Longrightarrow BX'\cdot BY=BI\cdot BL. \end{align*} And we obtain according to the item 1 that $IX'YL$ is inscribed quadrilateral.