Solution :
Since [tex]BE[/tex] is an altitude of a triangle [tex]ABC[/tex], it is perpendicular to [tex]AC[/tex] (since the triangle ).
Since [tex]G[/tex] and [tex]K[/tex] are the midpoints, then the line [tex]BG=GA[/tex] and the line [tex]BK=KC[/tex]. The line [tex]KG[/tex] must be parallel to the line [tex]AC[/tex] and is therefore perpendicular to [tex]BE.[/tex]
Now since F and [tex]G[/tex] are the mid points, we get [tex]GA=BG[/tex]. The line [tex]GF[/tex] is parallel to BH and is perpendicular to AC.
Therefore, we have
[tex]KG-GK[/tex] parallel to [tex]AC[/tex]
[tex]AC[/tex] perpendicular to [tex]BE-BH-BJ[/tex]
[tex]BE-BH-BJ[/tex] parallel to [tex]GF[/tex]
Thus [tex]GK[/tex] is perpendicular to [tex]GF[/tex] and also ∠[tex]FGK[/tex] is a right angle.
