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Math Help - Proving Collinearity

  1. #1
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    Proving Collinearity

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

    PQ and Q_1R_1 intersect at A.
    QR and P_1Q_1 intersect at B.

    Prove that the incentre, A, and B, are collinear.
    Last edited by Jameson; April 16th 2009 at 03:24 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 PQR.
    Let the bisector of each individual angle meet the circumcircle (for the second time). Let the points of intersection be P_1, Q_1, R_1.

    PQ and Q_1R_1 intersect at A.
    QR and P_1Q_1 intersect at B.

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

    It follows that P_1Q_1 bisects IR, and therefore the triangle BIR is isosceles. Hence the angle BIR is equal to \gamma. Show by similar arguments that the angle AIP is equal to \alpha, and conclude that the angle AIB is 180^\circ.
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    Quote Originally Posted by Opalg View Post
    \
    It follows that P_1Q_1 bisects 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 RR_1 meets P_1Q_1 at X. Then the triangles IXQ_1,\:RXQ_1 are congruent (because the two angles at Q_1 are both equal to \alpha, the angles at X are right angles, and the side XQ_1 is common to both triangles). Therefore IX=XR.
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