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$.

Bill Richardson 2010-11-12