1. ## 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.

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

3. Originally Posted by Opalg
\
It follows that $P_1Q_1$ bisects $IR$\
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 $RR_1$ 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 $IX=XR$.