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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 $EF$.

enter image description here

  • $\widehat{BIP}$ 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:

enter image description here

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.

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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:

enter image description here

  • 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 $OM$.

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