I was fiddling about on GeoGebra the other day when I noticed this:
In an acute-angled triangle ABC, let D, E and F be the points where the altitudes from A, B and C respectively meet the circumcircle of ABC again when extended.
The region where ABC and DEF overlap forms a hexagon. It turns out that the main diagonals of this hexagon (that is, the diagonals which each join two vertices that have exactly two other vertices in between them) are all concurrent at the orthocentre of ABC.
This is what I mean:
Let us label some points as follows:
We have put in: H, the orthocentre of ABC; W and X, where AB intersects CF and DF respectively; and Y and Z, where AC intersects BE and DE respectively. We will show that X, H and Z are collinear. Then we could apply this for the other diagonals.
Firstly, observe that angle CWB = angle BYC = 90; therefore, by the converse of angles in the same segment, BCYW is a cyclic quadrilateral.
Therefore, by angles in the same segment, angle WBY = angle WCY. (We could have obtained this also by applying angles in a triangle sum to 180 in triangle BYA, and then in BWC.)
But angle WCY is equal to angle FDA by angles in the same segment.
Furthermore, angle WBY is equal to angle ADE by angles in the same segment.
Hence, angle XBH = angle XDH; also angle HDZ = angle HCZ.
Thus, by the converse of angles in the same segment, BDHX and DCZH are cyclic quadrilaterals also.
What we have so far:
So, angle XBD = 180 – angle XHD (opposite angles in cyclic quadrilateral BDHX)
Angle ZCD = 180 – angle XBD = 180 – (180 – angle XHD) = angle XHD (opposite angles in cyclic quadrilateral ABDC)
Angle DHZ = 180 -angle ZCD = 180 – angle XHD (opposite angles in cyclic quadrilateral DCZH)
Therefore, since angle XHD and angle DHZ add up to 180, X, H and Z are collinear.
This completes the proof.
If you spot any mistakes let me know XD