Two Solutions

Exercise. If $AD, BE, CF$ are concurrent cevian lines for triangle $ABC$, show that $P(QR,AB)=-1.$

(Note. The following solutions require a knowledge of Ceva' and Menelaus' Theorems as well as knowledge of cross ratios, harmonic division and complete quadrilaterals.)

Solution 1. In the figure below extend $PQ$ to meet $AB$ in $Q'$. Let the cevian lines $AP, BQ, CR$ be concurrent on a point $X$.

Since $P$ and $Q$ are menelaus points, the point $Q'$ is such that $\dfrac{\overline{BR}}{\overline{RA}} = -\dfrac{\overline{BQ'}}{\overline{Q'A}}$, or $\dfrac{\overline{Q'A}}{\overline{BQ'}}\cdot\dfrac{\overline{BQ}}{\overline{QA}} = -1.$ Thus, $(Q'R,AB)=-1.$ Now $(Q'R,AB)=P(Q'R,AB)=P(QR,AB)$, and the result $P(QR,AB)=-1$ follows.

Solution 2. In the above figure, consider the quadrilateral $QCPX$. If we make this a complete quadrilateral, then $A$ and $B$ are vertices and $R$ and $Q'$ are diagonal points. We know that on the diagonal line $AB$ there is a harmonic range consisting of the two diagonal points and the two vertices lying on the diagonal line. Hence $(Q'R,AB)=-1$. But $(Q'R,AB)=P(Q'R,AB)=P(QR,AB)$ and it follows that $P(QR,AB)=-1$.