Show that $M$ is the midpoint of $JI$

Question

*$H$ is the orthocenter of $\triangle ABC$

I don't even know where to start, it appears that point $E$ has nothing too special about it, line $LI$ is too mysterious to me, I couldn't turn it into a radical axis or any convenient polar w.r.t some circle. Projective geometry also doesn't seem to help except if we could use Pascal's theorem in a tricky way, perhaps with the symedian of $\triangle ABC$ but I'm not sure how to do it.

EDIT: well, here is a transcription of the problem since people asked me: $\triangle JIK$ is the orthic triangle of $\triangle ABC$ and $L$ is the reflection of $H$ with respect to $BC$. Show that line $BE$ bisects $JI$ ($E = (ABC) \cap LI \neq L$).

Answer 1

Let us have an initial picture for the given problem:

The triangle $\Delta ABC$ is given, its orthocenter is $H$. We construct the ortic triangle $\Delta KIJ$. Let $L$ be the reflection of $H$ in $K$ (or w.r.t. $BC$). $LI$ intersects the circle for the second time in a point $E\ne L$. Let $M$ be the intersection $BE\cap IJ$.

Then: $M$ is the mid point of the segment $IJ$.

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Intermezzo: There is the notion of the cross ratio $r$ of four points on a line. It generalizes to the cross ratio $r_\circ$ of four points on a circle. As a reference i use Pamfilos' page Complex_Cross_Ratio.

If $A,B,C,D$ are points of a circle, which project from some $X$ on the same circle on $A',B',C',D'$ on some line, then $r_\circ(A,B;C,D)=r(A',B';C',D')$.

We use this property in the proof to conclude in a few lines.

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Proof: Consider the points $F,G$ on the circumcircle $\odot(ABC)$ so that:

• $F\ne B$ is the second intersection point of $BI$ with the circumicrcle,
• $G\ne B$ is taken so that $BG\|IJ$, so $\widehat{GLA}=\widehat{GBA}=\widehat{IJA}=\hat C$, and $\overset\frown{BA}=\overset\frown{AG}$.

We need below $BHIF\|LG$ and $IJ\|GB$ to read the projections of the infinity points.

Then: $$ \begin{aligned} \frac{MJ}{MI} &=r(M,\infty;J,I) \\ &=r_\circ(E,G;A,F) &&\text{ projecting from $B$ the circumcircle on $IJ$,} \\ &=r(I,\infty;H,F) &&\text{ projecting from $L$ the circumcircle on $BHIF$,} \\ &=\frac{IH}{IF}=1\ . \end{aligned} $$ Here, $\infty$ is in both cases the (one and the other) infinity point on the corresponding line involved in the cross ratio.

$\square$

Answer 2

Let's assume $\angle IBM =B_1$ and $\angle JBM= B_2.$

By the law of sines in $\triangle JIB,$ we have:

$$\frac{MI}{MJ}=\frac{BI}{BJ} \times \frac{\sin B_1}{\sin B_2}. \ \ (*)$$

In $\triangle BEI,$ we have:

$$\frac {\sin B_1}{\sin \angle BEL}=\frac{\sin B_1}{\cos B}=\frac{EI}{BI} \implies \sin B_1= \cos B \times \frac{EI}{BI}.$$

In $\triangle AEB,$ we have:

$$\frac {\sin B_2}{\sin \angle AEB}=\frac{\sin B_2}{\sin C}=\frac{AE}{AB} \implies \sin B_2= \sin C \times \frac{AE}{AB}.$$

Now, having $\sin B_1$ and $\sin B_2$ along with $(*)$, we get:

$$\frac{MI}{MJ}=\frac{\cos B}{\sin C} \times \frac{EI}{AE}\times \frac{AB}{BJ}. $$

On the other hand, $\triangle AEI$ and $\triangle ICL$ are similar. As a result, $\frac{EI}{AE}=\frac{IC}{LC}.$

Therefore, $$\frac{MI}{MJ}=\frac{\cos B}{\sin C} \times \frac{IC}{LC}\times \frac{AB}{BJ} \\ =\frac{\cos B}{\sin C}\times \frac {IC}{BJ} \times \frac{AB}{LC} \\ = \frac{\cos B}{\sin C} \times \frac{\cos C}{\cos B} \times \frac{\sin C}{\cos C}=1 \\ \implies MI=MJ. $$

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Remark: $\frac {IC}{BJ}=\frac{\cos C}{\cos B}$ and $\frac{AB}{LC}=\frac{\sin C}{\cos C}$ need to be verified, which is not a hard task.