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Math Help - Orthic triangle problem

  1. #1
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    Orthic triangle problem

    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 DD_{1}E = angle DD_{1}F = angle DD_{1}D_{2} 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.

    Thanks.
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  2. #2
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    Not sure I have understood but if "angle = angle = angle " These points form sets of angles then it seems F, E, D1, D2 can not be collinear.

    I may not have a correct understanding of "reflection" please clarify.
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  3. #3
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    BEFC is a cyclic quadrilateral. Therefore
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  4. #4
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    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
    Therefore
    angle EFO=angle OFD
    Now
    BF is parallel to DD_2
    Therefore
    angle OFD=angle FDD_2 = angle FD_2D
    Therefore
    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.
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  5. #5
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    Hello ordinalhigh

    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 E is the foot of the perpendicular from B to AC, and F the foot of the perpendicular from C to AB. (I think sa-ri-ga-ma has these points the wrong way round.) Suppose O is the orthocentre. (i.e the point where AD, BE and CF meet).


    Then:
    \angle BFC = \angle BEC = 90^o (given)

    \therefore BFEC is a cyclic quadrilateral (angles in same segment)


    \therefore \angle BEF = \angle BCF (same segment)

    Also:
    \angle ODC = \angle OEC = 90^o (given)

    \therefore OECD is a cylic quadrilateral (opposite angles supplementary)


    \angle BCF = \angle DCO (same angle)


    But \angle DCO = \angle DEO (same seg)
    =\angle DEB (same angle)
    But \angle BCF = \angle BEF (proven)

    \therefore \angle DEB = \angle BEF
    Now D_2 is the reflection of D in AC, and E lies on AC.
    \therefore DE = ED_2

    \therefore \angle ED_2D = \angle EDD_2 (isosceles \triangle EDD_2)
    =\angle DEB (alternate angles, BE \parallel DD_2)

    =\angle BEF (proven)

    \therefore D_2EF is a straight line (corresponding angles equal, BE \parallel DD_2)
    Similarly for D_1FE.

    Grandad
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