Etydhv

I've drawn the following diagram with Tikz:

The code used:

documentclassstandalone
usepackagepgf,tikz
usetikzlibrarybabel,calc,arrows,shapes.geometric,intersections,through,backgrounds

begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
% point O
coordinate (O) at (0,0);
% ABC triangle
node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
coordinate [label=above:$A$] (A) at (t.corner 1);
coordinate [label=left:$B$] (B) at (t.corner 2);
coordinate [label=right:$C$] (C) at (t.corner 3);
% O's label
node [above left] at (O) $O$;
% circle with 2.25cm radius and centre at O
draw (O) circle (2.25cm);
% point D: the point in the circumference whose angle is 50° with OC
coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
% radius OD
draw [name path=OD] (O) -- (D);
% point E: intersection between radius OD and the triangle
path [name intersections=of=OD and tri,by=E];
node [below] at (E) $E$;
% point F: point 33% the way from O to E
coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
% draw bullets at each point
foreach point in A,B,C,O,D,E,F
 fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument

About my current code:

• I'm aware of tkz-euclide and I think it could help me here, but the only documentation I could find (at CTAN) is in French, I language I can't read.


• According to the pgf documentation, the arrows library has been deprecated in favor of arrows.meta. I only used arrows because this code was adapted from Geogebra and I still haven't got into changing that particular aspect of the code.

I want to place 2 points G and H in the circle such that the chord GH:

• 
  passes through F and


• is parallel to AC

so as to obtain something like this (the following was drawn with Geogebra):

I've found answers regarding drawing parallel lines, but in this case it's not just parallel, it also needs to start and end at the circle, so I couldn't find a good answer by myself.

Define an (overlay such that it does not increase the bounding box) path that has the same slope as A--C (this is what let p1=($(C)-(A)$),n1=atan2(y1,x1) in does, which computes the angle of the line) and runs through F, and compute its intersections with the circle.

documentclass[tikz,border=3.14mm]standalone
usetikzlibraryarrows,calc,shapes.geometric,intersections
begindocument
begintikzpicture[line cap = round, line join = round, >=triangle 45, x=5.0cm, y=5.0cm]
 % point O
 coordinate (O) at (0,0);
 % ABC triangle
 node[name = t, name path = tri, regular polygon, regular polygon sides=3, minimum size=4.5cm, fill=lightgray!50, draw] at (O) ;
 coordinate [label=above:$A$] (A) at (t.corner 1);
 coordinate [label=left:$B$] (B) at (t.corner 2);
 coordinate [label=right:$C$] (C) at (t.corner 3);
 % O's label
 node [above left] at (O) $O$;
 % circle with 2.25cm radius and centre at O
 draw[name path=circle] (O) circle (2.25cm);
 % point D: the point in the circumference whose angle is 50° with OC
 coordinate [label=above right:$D$] (D) at ($(O)!1!50:(C)$);
 % radius OD
 draw [name path=OD] (O) -- (D);
 % point E: intersection between radius OD and the triangle
 path [name intersections=of=OD and tri,by=E];
 node [below] at (E) $E$;
 % point F: point 33% the way from O to E
 coordinate [label=above:$F$] (F) at ($(O)!.33!(E)$);
 path[overlay,name path=line] let p1=($(C)-(A)$),n1=atan2(y1,x1) in % computes the slope of A--C
 ($(F)+(n1:2*2.25cm)$) -- ($(F)+(180+n1:2*2.25cm)$);
 draw[name intersections=of=line and circle,by=G,H] (G) node[above left]$G$
 -- (H) node[below right]$H$;
 % draw bullets at each point
 foreach point in A,...,H
 fill [black] (point) circle (1.5pt);
endtikzpicture
enddocument

Alternatively you may just "parallel transport" the A--C path, which produces the same result and involves no atan2.

 path[overlay,name path=line]
 ($(F)+($(C)-(A)$)$) -- ($(F)+($(A)-(C)$)$);