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Thread: Proving Collinearity

  1. #1
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    Proving Collinearity

    Suppose we have an acute triangle $\displaystyle PQR$.
    Let the bisector of each individual angle meet the circumcircle (for the second time). Let the points of intersection be $\displaystyle P_1$, $\displaystyle Q_1$, $\displaystyle R_1$.

    $\displaystyle PQ$ and $\displaystyle Q_1R_1$ intersect at A.
    $\displaystyle QR$ and $\displaystyle P_1Q_1$ intersect at B.

    Prove that the incentre, A, and B, are collinear.
    Last edited by Jameson; Apr 16th 2009 at 02:24 AM. Reason: original question restored
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  2. #2
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    Quote Originally Posted by LegendWayne View Post
    Suppose we have an acute triangle $\displaystyle PQR$.
    Let the bisector of each individual angle meet the circumcircle (for the second time). Let the points of intersection be $\displaystyle P_1$, $\displaystyle Q_1$, $\displaystyle R_1$.

    $\displaystyle PQ$ and $\displaystyle Q_1R_1$ intersect at A.
    $\displaystyle QR$ and $\displaystyle P_1Q_1$ intersect at B.

    Prove that the incentre, A, and B, are collinear.
    Write $\displaystyle I$ for the incentre (it is the point where $\displaystyle PP_1,\: QQ_1$ and $\displaystyle RR_1$ all meet, of course). Write $\displaystyle 2\alpha,\:2\beta,\:2\gamma$ for the angles of the triangle, at $\displaystyle P,\:Q,\:R$ respectively. Now mark on a diagram all the angles you can find that are equal to $\displaystyle \alpha,\:\beta$ or $\displaystyle \gamma$. For example, the angles $\displaystyle QPP_1,\:RPP_1,\:QQ_1P_1,\:RQ_1P_1$ are all equal to $\displaystyle \alpha$. Also, $\displaystyle 2\alpha+2\beta+2\gamma=180^\circ$, and so $\displaystyle \alpha+\beta+\gamma=90^\circ$. Deduce that the lines $\displaystyle P_1Q_1$ and $\displaystyle RR_1$ are perpendicular.

    It follows that $\displaystyle P_1Q_1$ bisects $\displaystyle IR$, and therefore the triangle $\displaystyle BIR$ is isosceles. Hence the angle $\displaystyle BIR$ is equal to $\displaystyle \gamma$. Show by similar arguments that the angle $\displaystyle AIP$ is equal to $\displaystyle \alpha$, and conclude that the angle $\displaystyle AIB$ is $\displaystyle 180^\circ$.
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    Quote Originally Posted by Opalg View Post
    \
    It follows that $\displaystyle P_1Q_1$ bisects $\displaystyle IR$\
    It would be deeply appreciated if you can explain why it bisects.
    Thanks once again.
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  4. #4
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    Quote Originally Posted by LegendWayne View Post
    It would be deeply appreciated if you can explain why it bisects.
    Thanks once again.
    Suppose that $\displaystyle RR_1$ meets $\displaystyle P_1Q_1$ at $\displaystyle X$. Then the triangles $\displaystyle IXQ_1,\:RXQ_1$ are congruent (because the two angles at $\displaystyle Q_1$ are both equal to $\displaystyle \alpha$, the angles at $\displaystyle X$ are right angles, and the side $\displaystyle XQ_1$ is common to both triangles). Therefore $\displaystyle IX=XR$.
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