Question

Let ABC be a triangle. Let E be a point on the segment BC such that $BE=2EC$. Let F be the mid-point of AC. Let BF intersect AE in Q. Determine $BQ=QF$.

Answer 1

Let CQ and ET meet AB in S and T respectively. We have
$\frac{\left[SBC\right]}{\left[ASC\right]}=\frac{BS}{SA}=\frac{\left[SBQ\right]}{\left[ASQ\right]}$
Using componendo by dividendo, we obtain
$\frac{BS}{SA}=\frac{\left[SBC\right]-\left[SBQ\right]}{\left[ASC\right]-\left[ASQ\right]}=\frac{\left[BQC\right]}{\left[AQC\right]}$
Similarly, We can prove
$\frac{BE}{EC}=\frac{\left[BQA\right]}{\left[CQA\right]},\frac{CF}{FA}=\frac{\left[CQB\right]}{\left[AQB\right]}$
But BD = DE = EC implies that BE / EC = 2; CF = FA gives CF / FA = 1. Thus
$\frac{BS}{SA}=\frac{\left[BQC\right]}{\left[AQC\right]}=\frac{\left[BQC\right]/\left[AQB\right]}{\left[AQC\right]/\left[AQB\right]}=\frac{CF/FA}{EC/BE}=\frac{1}{1/2}=2$.
Now
$\frac{BQ}{QF}=\frac{\left[BQC\right]}{\left[FQC\right]}=\frac{\left[BQA\right]}{\left[FQA\right]}=\frac{\left[BQC\right]+\left[BQA\right]}{\left[FQC\right]+\left[FQA\right]}=\frac{\left[BQC\right]+\left[BQA\right]}{\left[AQC\right]}$
This gives
$\frac{BQ}{QF}=\frac{\left[BQC\right]=\left[BQA\right]}{\left[AQC\right]}=\frac{\left[BQC\right]}{\left[AQC\right]}+\frac{\left[BQA\right]}{\left[AQC\right]}+\frac{\left[BQA\right]}{\left[AQC\right]}=\frac{BS}{SA}+\frac{BE}{EC}=2+2=4$
[Note : BS / SA can also be obtained using Cevas theorem. One can also obtain the result by coordinate geometry.)