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 Q respectively. Determine the ratio of the area of the triangle APQ to that of the quadrilateral PDEQ.

Answer 1

If we can find [APQ] / ADE], then we can get the required ratio as
$\frac{\left[APQ\right]}{\left[PDFQ\right]}=\frac{\left[APQ\right]}{\left[ADE\right]-\left[APQ\right]}=\frac{1}{\left(\left[ADE\right]/\left[APQ\right]\right)-1}$
Now draw $PM\perp AEandDL\perp AE$. Observe
$\left[APQ\right]=\frac{1}{2}AQ.PM,\left[ADE\right]=\frac{1}{2}AE.DL$
Further, since $PM\parallel DL$, we also get PM/DL = AP/AD. Using these we obtain
$\frac{\left[APQ\right]}{\left[ADE\right]}=\frac{AP}{AD}.\frac{AQ}{AE}$.
We have
$\frac{AQ}{QE}=\frac{\left[ABQ\right]}{\left[ECQ\right]}=\frac{\left[ACQ\right]}{\left[ECQ\right]}=\frac{\left[ABQ\right]+\left[ACQ\right]}{\left[BCQ\right]}=\frac{\left[ABQ\right]}{\left[BCQ\right]}+\frac{\left[ACQ\right]}{\left[BCQ\right]}=\frac{AF}{FC}=\frac{AS}{SB}$.
However
$\frac{BS}{SA}=\frac{\left[BQC\right]}{\left[AQC\right]}=\frac{\left[BQC\right]/\left[AQB\right]}{\left[AQC\right]/\left[AQB\right]}=\frac{CF/FA}{EC/BE}=\frac{1}{1/2}=2$
Besides AF/FC = 1. We obtain
$\frac{AQ}{QE}=\frac{AF}{FC}+\frac{AS}{SB}=1+\frac{1}{2}=\frac{3}{2},\frac{AE}{QE}=1+\frac{3}{2}=\frac{5}{2},\frac{AQ}{AE}=\frac{3}{5}$
Since $EF\parallel AD$ (since DE/EC = AF/FC = 1), we get AD = 2 EF. Since $EF\parallel PD$, we also have PD/EF = BD/DE = 1/2. Hence EF = 2 PD. Thus AD = 4 PD. This gives and AP/PD = 3 and AP/AD = 3/4. Thus
$\frac{\left[APQ\right]}{\left[ADE\right]}=\frac{AP}{AD}.\frac{AQ}{AE}=\frac{3}{4}.\frac{3}{5}=\frac{9}{20}$.
Finally,
$\frac{\left[APQ\right]}{\left[PDEQ\right]}=\frac{1}{\left(\left[ADE\right]/\left[APQ\right]\right)-1}=\frac{1}{\left(20/9\right)-1}=\frac{9}{11}$.
(Note: BS/SA can also be obtained using Cevas theorem. Coordinate geometrysolution can also be obtained.)