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Thread: Orthic Triangle and Angle Bisector

  1. #1
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    Orthic Triangle and Angle Bisector

    Firstly, it's really hard to enter a "descriptive title" for this, but that's beside the point

    Anyways, here's the problem I have:

    If $\displaystyle PP'$, $\displaystyle QQ'$ and $\displaystyle RR'$ are altitudes of triangle $\displaystyle PQR$, and $\displaystyle X$ and $\displaystyle Y$ are points on $\displaystyle P'R'$ and $\displaystyle Q'R'$ respectively such that $\displaystyle \angle XPY=\angle P'PR$, prove that $\displaystyle PX$ bisects $\displaystyle \angle R'XY$.

    I've done substantial angle chasing with the orthic triangle, but I can't seem to prove that it is an angle bisector. I've used only 2 pronumerals for angles: $\displaystyle \alpha$ for $\displaystyle \angle P'PR$ and $\displaystyle \theta$ for $\displaystyle \angle R'XP$ and I'm trying to prove $\displaystyle \angle PXY$ is also $\displaystyle \theta$. Is this the correct/ a good approach? If not, what'd be better? Some hints would do fine, all I've got atm is a huge bunch of angle chasing rubbish
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  2. #2
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    Dec 2008
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    Ah well, never mind, I finally found a solution
    The trick was finding the millions of cyclic quads (ok, slightly exaggerated...)

    I can post my sol'n up here if anyone wants it, but otherwise, see ya guys later
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