About a medial line, incircle, and circumcircle a triangle

Question

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:

Answer 1

I propose another solution. Let's first state two facts that we will use in the proof.

1. 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 $$ KA_1\cdot KA_2=KB_1\cdot KB_2. $$

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.

3. 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.

Answer 2

Consider the figure below. First, we state a lemma.

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Lemma: $BI^2=2.IK.BD.$

Proof: I don't write a full proof; however this is just a problem of expressing all the variables in terms of the angles and a common edge after which the lemma boils down to verifying a trigonometric identity. Indeed, we have: $$BI=AB \frac{\sin \frac{A}{2}}{\sin (\frac{B}{2}+\frac{A}{2})}, \\IK=BI-\frac{1}{2}AB \frac{\sin A}{\sin (\frac{B}{2}+A)}, \\ BD=AB \frac{\sin (A+\frac{B}{2})}{\sin C}.$$

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Step $1$: Let's assume $\angle LXD=a, \angle KYI=b$, and $\angle XDB=\angle XYB=c$.

Then; $$\frac{\frac{\sin a}{\sin (a+c)}}{\frac{\sin b}{\sin (b+c)}}=\frac{\frac{DL}{XD}}{\frac{\frac{\sin b}{\sin \angle KIY}}{\frac{\sin (b+c)}{\sin \angle KIY}}}=\frac{DL}{XD}.\frac{KY}{KI}.\frac{BI}{BY}.$$

But it is very easy (you can use the law of sine) to show that: $$\frac{KY}{BY}=\frac{XD}{BD}.$$

Now, by the Lemma, we conclude that:

$$\frac{\frac{\sin a}{\sin (a+c)}}{\frac{\sin b}{\sin (b+c)}}=1 \implies \frac{\sin (b+c)}{\sin b}=\frac{\sin (a+c)}{\sin a} \\ \implies \cos c+\cot b \sin c= \cos c+\cot a \sin c \implies a=b.$$

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Step $2$: Similarly, this time let's assume $\angle LYD=a, \angle KXI=b$, and $\angle YDB=\angle BXY=c$. A very similar argument as that of step $1$ (the argument is almost the same and just $X$ and $Y$ are replaced in the fractions), proves that $a=b.$

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Step $3$: So far, We have proved that:

$$\angle LXD=\angle KYI, \\ \angle LYD=\angle KXI.$$

Moreover note that:

$$\angle KXD=\angle KYD,$$

from which the claim follows.

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We are done.