1. ## 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.

2. Originally Posted by LegendWayne
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 $\displaystyle 180^\circ$.

3. Originally Posted by Opalg
\
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.

4. Originally Posted by LegendWayne
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 $\displaystyle IX=XR$.