Suppose that triangle ABC is an acute-angle triangle, D is the foot of the altitude from A and D1 and D2 are the reflections of D with respect to AB and AC, respectively. Let F and E be the feet of the altitudes from C and B respectively. Show that the points F, E, D1, D2 are collinear.
I am thinking that if I can show that angle = angle = angle then that would show the points are collinear. I had a thought I can establish this using cyclic quadrilaterals but I got stuck. I can use some help on what to do next.
BEFC is a cyclic quadrilateral. Therefore
angle BFE = angle BCE
If O is the orthocenter, then DOFC is a cyclic quadrilateral. Therefore
angle OFD=angle OCD
angle EFO=angle OFD
BF is parallel to DD_2
angle OFD=angle FDD_2 = angle FD_2D
angle EFD=angle FDD_2 + angle FD_2D
Hence EFD_2 is a straight line.
Similarly you can show D_1EF also a straight line.
sa-ri-ga-ma has given you most of the proof, but if I may, I'd like to make one or two corrections and additions.
First, note that is the foot of the perpendicular from to , and the foot of the perpendicular from to . (I think sa-ri-ga-ma has these points the wrong way round.) Suppose is the orthocentre. (i.e the point where and meet).
is a cyclic quadrilateral (angles in same segment)
(given)Now is the reflection of in , and lies on .
is a cylic quadrilateral (opposite angles supplementary)
But (same seg)(same angle)But (proven)
Similarly for .
(isosceles )(alternate angles, )is a straight line (corresponding angles equal, )