A line that goes through the orthocenter and parallel to one side of the triangle

Given an acute $\Delta ABC$ $(AB<AC)$ with circumcenter $O$ and orthocenter $H$.

• Line $AO$ intersects $BC$ at $D$.

• Line segment $AC$ intersects the circumcircle of $\Delta ABD$ at point $E$.

• Line segment $AB$ intersects the circumcircle of $\Delta ACD$ at point $F$.

• Line segments $BE$ and $CF$ intersects at $K$.

Prove that $HK$ is parallel to $BC$.

I actually knew how to prove this and succesfully proved it, but I want to know if my proof is correct and if there is another better way, so I posted both the figure and my attempt as my own answer.

• Sorry I mistyped it, the suggested edit is correct. – user061703 Mar 14 '18 at 14:22

From the brown circle and then the blue circle, we have $\angle BHC = \angle BKC$ because they both are equal to $180^0 - \angle BAC$. This further means BHKC is cyclic.

Then, apply your finding (like HI = KI and BI = CI etc.) to get the required result.

• Let $AA'$ be the diameter of the circle $(O)$. Draw two altitudes of $\Delta ABC$, $BE'$ and $CF'$. $CK$ and $BH$ intersect at $L$; $CH$ and $BK$ intersect at $I$.

• $ABDE,AFDC,ABA'C$ are cyclic quadrilaterals, which implies $\widehat{EBC}=\widehat{A'AC}=\widehat{A'BC}$ and $\widehat{FCB}=\widehat{BAA'}=\widehat{BCA'}$ or $\widehat{KBC}=\widehat{A'BC}$ and $\widehat{KCB}=\widehat{A'CB}$.

• Apply the Thales's theorem, we will have $\widehat{ABA'}=\widehat{ACA'}=90^\circ$ for right triangles $\Delta ABA'$ and $\Delta ACA'$.

• We have drawn extra altitudes for the triangle, which implies that $CF'$ is parallel to $BA'$ and $BE'$ is parallel to $CA'$. Combined with the second step, we will have $\widehat{HCB}=\widehat{CBA'}=\widehat{KBC}$ and $\widehat{E'BC}=\widehat{BCA'}=\widehat{BCK}$.

• The step above implies $\Delta{IBC}$ isosceles at $I$ and $\Delta{LBC}$ isosceles at $L$, which implies $IB=IC$ and $\widehat{HBI}=\widehat{LBC}-\widehat{IBC}=\widehat{LCB}-\widehat{ICB}=\widehat{LCI}$.

• $\Delta{IHB}$ and $\Delta{IKC}$ are congruent (ASA: $IB=IC;\widehat{HBI}=\widehat{KCI};\widehat{HIB}=\widehat{KIC}$), so we have $IH=IK$ and $IB=IC$, implies $\frac{IH}{IC}=\frac{IK}{IB}$, but we also have $\widehat{HIK}=\widehat{BIC}$, so $\Delta{IHK}$ and $\Delta{ICB}$ are similar $\Rightarrow \widehat{IHK}=\widehat{ICB} \Rightarrow HK$ is parallel to $BC$.