Question

Let ABC be a triangle. Let D, E be a points on the segment BC such that BD = DE = EC. Let F be the mid-point of AC. Let BF intersect AD in P and AE in Qrespectively. Determine BP / PQ.

Answer 1

Let D be the mid-point of BE.Join AD and let it intersect BF in P. Extend CQ and EP to meet AB in S and T respectively. Now
$\frac{BS}{SA}=\frac{\left[BQC\right]}{\left[AQC\right]}=\frac{\left[BQC\right]/\left[AQB\right]}{\left[AQC\right]/\left[AQB\right]}=\frac{CF/FA}{EC/BE}=\frac{1}{1/2}=2$
Similarly,
$\frac{AQ}{QE}=\frac{\left[ABQ\right]}{\left[EBQ\right]}=\frac{\left[ACQ\right]}{\left[ECQ\right]}=\frac{\left[ABQ\right]+\left[ACQ\right]}{\left[BCQ\right]}=\frac{\left[ABQ\right]}{\left[BCQ\right]}+\frac{\left[ACQ\right]}{\left[BCQ\right]}=\frac{AF}{FC}+\frac{AS}{SB}=1+\frac{1}{2}=\frac{3}{2}$
And
$\frac{AT}{TB}=\frac{\left[APE\right]}{\left[BPE\right]}=\frac{\left[APE\right]}{\left[APB\right]}\cdot \frac{\left[APB\right]}{\left[BPE\right]}=\frac{DE}{DB}\cdot \frac{AQ}{QE}=1\cdot \frac{3}{2}=\frac{3}{2}$
Finally,
$\frac{BP}{PQ}=\frac{\left[BPE\right]}{\left[QPE\right]}=\frac{\left[BPA\right]}{\left[APE\right]}=\frac{\left[BPQ\right]+\left[BPA\right]}{\left[APE\right]}=\frac{\left[BPE\right]}{\left[APE\right]}+\frac{\left[BPA\right]}{APE}=\frac{BT}{TA}=\frac{BD}{DE}=\frac{2}{3}+1=\frac{5}{3}$
(Note: BS / SA, AT / TB can also be obtained using Cevas theorem. A solution can also be obtained using coordinate geometry.)