proving lines are perpendicular Let $BCD$ $(BC<BD)$ be a triangle inscribed in circle $(I)$. Let $E\in BC, F\in BD, G\in CD$ be such that $DE\perp BC, CF\perp BD, BG\perp CD$. Let $L$ be the orthocenter of triangle $BCD$. Let $H$ be the intersection point of $EG$ and $CF$, $J$ be the intersection point of $GF$ and $DE$. Let $K$ be the midpoint of $LI$.
Prove that $BK\perp HJ$.

 A: This is NOT a solution. I just want to share my finding which is too long to be included in a comment.

As pointed out, K is the center of the nine-point circle. Its properties, together with LK = KI, generate several pairs of parallel lines (marked in blue) but they don’t seem to help.
The standard way to prove $BK \bot HJ$ is to prove $\angle BMH = 90^0$; where M is the intersection of BK and HJ. This is done if we can prove BEBJ is cyclic by showing $\angle MJE = \angle MBE$. Or alternately, if we can show the purple lines are parallel; where EN is drawn to be perpendicular to BK extended.
A: I am gonna continue on Mick's writings and image. I hope you know using trigonometry on geometry.
We should prove  $\angle MJE = \angle MBE$ , so $\angle HJL = \angle KBE$. $$ $$
We know that $\angle HJL+\angle JHL=B=\angle KBE+\angle KBF$
$$ $$
So it's enough to show that  $$\dfrac {\sin(HJL)}{\sin(JHL)}=\dfrac {\sin(KBE)}{\sin(KBF)}$$
$$ $$ (Left as an exercise: if $x+y=z+t\lt180$ and $\dfrac {\sin(x)}{\sin(y)}=\dfrac {\sin(z)}{\sin(t)}$, then $x=z$ and $y=t$)
$$ $$
Let $P$ and $Q$ be the projections of $K$ onto $BC$ and $BD$, respectively. Then,
$KP=KB*\sin(KBE)$ and $KQ=KB*\sin(KBF)$, so 
$$\dfrac {\sin(KBE)}{\sin(KBF)}=\dfrac {KP}{KQ}$$
and also (Let $R$ be the radius of circumcircle of $BCD$)$$KP=\dfrac {LE+IT}{2}=\dfrac {2R\cos(B)\cos(C)+R\cos(D)}{2}=\dfrac {R\cos(B-C)}{2}$$,
$$KQ=\dfrac {LF+IS}{2}=\dfrac {2R\cos(B)\cos(D)+R\cos(C)}{2}=\dfrac {R\cos(B-D)}{2}$$
$$\Rightarrow \dfrac {\sin(KBE)}{\sin(KBF)}=\dfrac {KP}{KQ}=\dfrac {\cos(B-C)}{\cos(B-D)}\tag {1}$$
$$ $$
By the triangle $HJL$,
$$\dfrac {\sin(HJL)}{\sin(JHL)}=\dfrac {HL}{JL}=\dfrac {HL}{EL}*\dfrac {EL}{JL}$$
Because $\angle HEL=90-D$ and $\angle EHL=90+D-B$  $\quad$(easy to see)
$$\dfrac {HL}{EL}=\dfrac {\cos(D)}{\cos(B-D)}$$
Because $GL$ is bisecting $\angle EGJ$
and because  $\angle GEJ=90-D$ and $\angle GJE=90+B-C$  $\quad$(easy to see)
$$\dfrac {EL}{JL}=\dfrac {EG}{GJ}=\dfrac {\cos(B-C)}{\cos(D)}$$
$$\Rightarrow \dfrac {\sin(HJL)}{\sin(JHL)}=\dfrac {HL}{EL}*\dfrac {EL}{JL}=\dfrac {\cos(B-C)}{\cos(B-D)}\tag {2}$$
$$ $$
By (1) and (2),
$$\dfrac {\sin(HJL)}{\sin(JHL)}=\dfrac {\cos(B-C)}{\cos(B-D)}=\dfrac {\sin(KBE)}{\sin(KBF)}$$
$$ $$
$$\Rightarrow \angle HJL = \angle KBE $$
 $$\Rightarrow \angle MJE = \angle MBE $$
