
5. Let $A B C$ be a triangle. Let $D, E$ be a points on the segment $B C$ such that $B D=$ $D E=E C$. Let $F$ be the mid-point of $A C$. Let $B F$ intersect $A D$ in $P$ and $A E$ in $Q$ respectively. Determine $B P / P Q$.


Solution: Let $D$ be the mid-point of $B E$.

![](https://cdn.mathpix.com/cropped/2024_06_05_bf025a87579e845b659fg-2.jpg?height=331&width=545&top_left_y=699&top_left_x=454)
Join $A D$ and let it intersect $B F$ in $P$. Extend $C Q$ and $E P$ to meet $A B$ in $S$ and $T$ respectively. Now

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

Similarly,

$$
\frac{A Q}{Q E}=\frac{[A B Q]}{[E B Q]}=\frac{[A C Q]}{[E C Q]}=\frac{[A B Q]+[A C Q]}{[B C Q]}=\frac{[A B Q]}{[B C Q]}+\frac{[A C Q]}{[B C Q]}=\frac{A F}{F C}+\frac{A S}{S B}=1+\frac{1}{2}=\frac{3}{2}
$$

And

$$
\frac{A T}{T B}=\frac{[A P E]}{[B P E]}=\frac{[A P E]}{[A P B]} \cdot \frac{[A P B]}{[B P E]}=\frac{D E}{D B} \cdot \frac{A Q}{Q E}=1 \cdot \frac{3}{2}=\frac{3}{2}
$$

Finally,

$$
\frac{B P}{P Q}=\frac{[B P E]}{[Q P E]}=\frac{[B P A]}{[A P E]}=\frac{[B P E]+[B P A]}{[A P E]}=\frac{[B P E]}{[A P E]}+\frac{[B P A]}{[A P E]}=\frac{B T}{T A}+\frac{B D}{D E}=\frac{2}{3}+1=\frac{5}{3}
$$

(Note: $B S / S A, A T / T B$ can also be obtained using Ceva's theorem. A solution can also be obtained using coordinate geometry.)
