
5. Let $A B C$ be a triangle. Let $E$ be a point on the segment $B C$ such that $B E=2 E C$. Let $F$ be the mid-point of $A C$. Let $B F$ intersect $A E$ in $Q$. Determine $B Q / Q F$.


Solution: Let $C Q$ and $E T$ meet $A B$ in $S$

![](https://cdn.mathpix.com/cropped/2024_06_05_2df523a359bae2f5e442g-2.jpg?height=328&width=545&top_left_y=671&top_left_x=454)
and $T$ respectively. We have

$$
\frac{[S B C]}{[A S C]}=\frac{B S}{S A}=\frac{[S B Q]}{[A S Q]}
$$

Using componendo by dividendo, we obtain

$$
\frac{B S}{S A}=\frac{[S B C]-[S B Q]}{[A S C]-[A S Q]}=\frac{[B Q C]}{[A Q C]}
$$

Similarly, We can prove

$$
\frac{B E}{E C}=\frac{[B Q A]}{[C Q A]}, \quad \frac{C F}{F A}=\frac{[C Q B]}{[A Q B]}
$$

But $B D=D E=E C$ implies that $B E / E C=2 ; C F=F A$ gives $C F / F A=1$. Thus

$$
\frac{B S}{S A}=\frac{[B Q C]}{[A Q C]}=\frac{[B Q C] /[A Q B]}{[A Q C] /[A Q B]}=\frac{C F / F A}{E C / B E}=\frac{1}{1 / 2}=2
$$

Now

$$
\frac{B Q}{Q F}=\frac{[B Q C]}{[F Q C]}=\frac{[B Q A]}{[F Q A]}=\frac{[B Q C]+[B Q A]}{[F Q C]+[F Q A]}=\frac{[B Q C]+[B Q A]}{[A Q C]}
$$

This gives

$$
\frac{B Q}{Q F}=\frac{[B Q C]+[B Q A]}{[A Q C]}=\frac{[B Q C]}{[A Q C]}+\frac{[B Q A]}{[A Q C]}=\frac{B S}{S A}+\frac{B E}{E C}=2+2=4
$$

(Note: $B S / S A$ can also be obtained using Ceva's theorem. One can also obtain the result by coordinate geometry.)
