Consider $(O, r)$ being the circumcircle of $\triangle ABC$ where $BC$ is fixed and $A$ is not. $OE$ and $OF$ are perpendicular to respectively $AB$ and $AC$ ($E \in AC$, $F \in AB$). $(E, EA) \cap (F, FA) = K$ ($K \not\equiv A$). Prove that $K$ lies on a fixed circle and $AK$ passes through a fixed point where $\dfrac{BC}{r} = \sqrt 3$.
As can be seen, $K \in (O, B, C)$ and $AK$ passes through $O$. But I can't find out a method to the second one.
The first can be solved like so.
$\widehat{BOC} = 2 \cdot \sin^{-1}\dfrac{BC}{r} = 120^\circ \implies \widehat{CAB} = 60^\circ$.
We have that $\widehat{BKC} = \widehat{BKA} + \widehat{CKA} = \dfrac{\widehat{BEA}}{2} + \dfrac{\widehat{CFA}}{2} = 90^\circ - \widehat{BAE} + 90^\circ - \widehat{CAF}$
$180^\circ - 2 \cdot \widehat{CAB} = 180^\circ - 2 \cdot 60^\circ = 60^\circ$ $\implies \widehat{BOC} + \widehat{BKC} = 120^\circ + 60^\circ = 180^\circ$.
That means $K \in (O, B, C)$.
I think there is no necessary for given numbers.
first ,to prove $K$ is on the fixed circle: $EO \perp AB \implies AE=BE \implies B \in circle (E,AE)$,
same reason,$C \in circle(F,AF)$ ,connect $KB,KC,LC,MB$ $\angle CKA=\angle CLA, \angle BKA= \angle BMA$
$AL$ is diameter $\implies \angle CLA= 90^\circ- \angle BAC$
$AM$ is diameter $\implies \angle BMA= 90^\circ- \angle BAC$
$\implies \angle CKA=\angle BKA , \angle BAC= \dfrac{\angle BOC}{2} \implies \angle CKB+\angle BOC=180^\circ \implies K \in circle (O,B,K,C) $
2nd, to prove AK pass a fixed point:
let $AK$ cross $circle(O,B,K,C)$ at point $O'$, since $\angle CKA=\angle BKA \implies O'B=O'C$
$BO=CO \implies O'=O \implies AK $ pass $O$
3rd to prove $F \in circle(O,B,K,C)$ :
$\angle OFA + \angle BAC =90^\circ \implies \angle OCB=\angle OFA \implies \angle OCB+\angle OFB =180^\circ \implies F \in circle(O,B,K,C)$
4th to prove $N \in circle(A,B,C)$ :
$\angle BNL + \angle BLN =90^\circ \implies \angle BNL=\angle BAC \implies \angle BNC + \angle BAC=180^\circ \implies N \in circle(A,B,C)$