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Thread: 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 $\displaystyle DD_{1}E$ = angle $\displaystyle DD_{1}F$ = angle $\displaystyle DD_{1}D_{2}$ 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 $\displaystyle E$ is the foot of the perpendicular from $\displaystyle B$ to $\displaystyle AC$, and $\displaystyle F$ the foot of the perpendicular from $\displaystyle C$ to $\displaystyle AB$. (I think sa-ri-ga-ma has these points the wrong way round.) Suppose $\displaystyle O$ is the orthocentre. (i.e the point where $\displaystyle AD, BE$ and $\displaystyle CF$ meet).


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

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


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

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

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


    $\displaystyle \angle BCF = \angle DCO$ (same angle)


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

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

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

    $\displaystyle =\angle BEF$ (proven)

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

    Grandad
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