# Parallel property of $LK$ and $OM$ where $L$, $M$ are points on the median $AM$ of $\Delta{ABC}$

Given an acute $\Delta{ABC}$ $(AB<AC)$ with the circumcenter $O$, two altitudes $BE, CF$ and the median $AM$. Let $IB$ and $IC$ be the two tangents of $(O)$ so that $AI$ intersects $BC$ at $K$. Segment $EF$ intersects the median $AM$ at $L$. Prove that $LK$ is parallel to $OM$.

My attempt:

Lemma $1$: If $AI$ intersects $EF$ at $M'$, prove that $M'$ is the midpoint of $EF$. • $\widehat{BIP}$ is formed by the chord $BP$ and the tangent $BI$, while $\widehat{BAP}$ is an inscribed angle. Similarly, $\widehat{CIP}$ is formed by the chord $BP$ and the tangent $BI$, while $\widehat{CAP}$ is an inscribed angle. Because of this, $\widehat{IBP}=\widehat{IAB}$ and $\widehat{ICP}=\widehat{IAC}$, implying $\Delta{IPB}$ is similar to $\Delta{IBA}$ and $\Delta{IPC}$ is similar to $\Delta{ICA}$.

• Two pairs of similar triangles above imply that $\dfrac{IP}{IB}=\dfrac{PB}{BA}$ and $\dfrac{IP}{IC}=\dfrac{PC}{CA}$. However, $IB=IC$ for both being tangents of $(O)$, so we now know that $\dfrac{PB}{BA}=\dfrac{PC}{CA}\Rightarrow\dfrac{PB}{PC}=\dfrac{AB}{AC}$.

• Because $BFEC$ is a cyclic quadrilateral $(\widehat{BFC}=\widehat{BEC}=90^\circ)$, we can prove that $\Delta{AEF}$ is similar to $\Delta{ABC}$, so $\dfrac{AE}{AF}=\dfrac{AB}{AC}=\dfrac{PB}{PC}\Rightarrow AF.BP=AE.CP$

• Also because $BFEC$ is a cyclic quadrilateral and the inscribed angle theorem, we know that $\widehat{AFM'}=\widehat{ACB}=\widehat{APB}$ and $\widehat{AEM'}=\widehat{ABC}=\widehat{APC}$, so $\Delta{AM'F}$ is similar to $\Delta{ABP}$, $\Delta{AM'E}$ is similar to $\Delta{ACP}$, so $\dfrac{M'F}{BP}=\dfrac{AF}{AP}$ and $\dfrac{M'E}{CP}=\dfrac{AE}{AP}\Rightarrow M'F.AP=AF.BP$ and $M'E.AP=AE.CP$.

• ${\begin{cases}AF.BP=AE.CP\\M'F.AP=AF.BP\\M'E.AP=AE.CP\end{cases}\Rightarrow }$ $M'$ is the midpoint of $EF$.

I don't know if the first lemma has any relation to the actual problem, but here is another lemma that I'm wokring on:

Lemma $2$: The line that goes through $K$ and perpendicular to the line $BC$ intersects $FE$ at $L'$. From $L'$, draw a segment $GH$ goes through $L'$ and parallel to the line $BC$. Prove that $L'$ is the midpoint of $GH$.

Because I have already proven that $M'$ is the midpoint of $EF$, I can now get rid of the point $I$ to make the figure easier: At this point I'm stuck, as I'm unable to proof that $L'$ is the midpoint of $GM$ or $\Delta{KGH}$ is an isoceles triangle or $KG=KH$.

If we can prove that $L'$ is the midpoint of $GH$, then combine with the intercept theorem, then $AL'$ goes through the midpoint of $BC$, or $A,L',M$ are collinear, then $L$ and $L'$ are two same points because they both lie on $EF$, then because $KL'$ is perpendicular to $BC$, $KL$ is also perpendicular to $BC$. Finally, we can prove that because $OM$ is perpendicular to $BC$ so it is parallel to $KL$.

However, the last part still needs $L'$ to be the midpoint of $GH$, hope someone can help me finish this proof, based on all the information above.

Thank you for your interest in this problem, but after thinking more about this problem for two days, I have managed to solve it myself.

I will bring back the figure again: • For the lemma $1$ (see question), I have already proven that $M'$ is the midpoint of $EF$.

• $\Delta{FBC}$ has the median $FM$, $\Delta{EBC}$ has the median $EM$ and $\widehat{BFC}=\widehat{BEC}=90^\circ$ so $MF=ME=\dfrac{BC}{2}$, implying that the isoceles $\Delta{MEF}$ has the median $MM'$ so it is also the altitude of the triangle, or $\widehat{MM'L}=90^\circ$.

• We have already proven that $\Delta{AEF}$ is similar to $\Delta{ABC}$, so $\dfrac{AF}{AC}=\dfrac{FE}{BC}=\dfrac{2FM'}{2MC}=\dfrac{FM'}{MC}$ because $M,M'$ are the respective midpoints of of $BC$ and $EF$. We also have $\widehat{AFM'}=\widehat{ACM}$ so $\Delta{AM'F}$ is similar to $\Delta{AMC}$.

• Because $\Delta{AM'F}$ is similar to $\Delta{AMC}$, $\widehat{AMC}=\widehat{AM'F}=\widehat{LM'K}$ (vertical angles). Because of this, $LM'KM$ is a cyclic quadrilateral. This means $\widehat{LKM}=\widehat{LM'M}=90^\circ=\widehat{OMK}$ or $LK$ is parallel to $OM$.