
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 the ratio of the area of the triangle $A P Q$ to that of the quadrilateral $P D E Q$.


Solution: If we can find $[A P Q] /[A D E]$, then we can get the required ratio as

![](https://cdn.mathpix.com/cropped/2024_06_05_c2bde1f57585502709b7g-2.jpg?height=331&width=547&top_left_y=794&top_left_x=450)

$$
\begin{aligned}
\frac{[A P Q]}{[P D E Q]}=\frac{[A P Q]}{[A D E]-[A P Q]} \\
\quad=\frac{1}{([A D E] /[A P Q])-1}
\end{aligned}
$$

Now draw $P M \perp A E$ and $D L \perp A E$. Observe

$$
[A P Q]=\frac{1}{2} A Q \cdot P M,[A D E]=\frac{1}{2} A E \cdot D L
$$

Further, since $P M \| D L$, we also get $P M / D L=A P / A D$. Using these we obtain

$$
\frac{[A P Q]}{[A D E]}=\frac{A P}{A D} \cdot \frac{A Q}{A E}
$$

We have

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

However

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

Besides $A F / F C=1$. We obtain

$$
\frac{A Q}{Q E}=\frac{A F}{F C}+\frac{A S}{S B}=1+\frac{1}{2}=\frac{3}{2}, \quad \frac{A E}{Q E}=1+\frac{3}{2}=\frac{5}{2}, \quad \frac{A Q}{A E}=\frac{3}{5}
$$

Since $E F \| A D$ (since $D E / E C=A F / F C=1$ ), we get $A D=2 E F$. Since $E F \| P D$, we also have $P D / E F=B D / D E=1 / 2$. Hence $E F=2 P D$. Thus $A D=4 P D$. This gives and $A P / P D=3$ and $A P / A D=3 / 4$. Thus

$$
\frac{[A P Q]}{[A D E]}=\frac{A P}{A D} \cdot \frac{A Q}{A E}=\frac{3}{4} \cdot \frac{3}{5}=\frac{9}{20}
$$

Finally,

$$
\frac{[A P Q]}{[P D E Q]}=\frac{1}{([A D E] /[A P Q])-1}=\frac{1}{(20 / 9)-1}=\frac{9}{11}
$$

(Note: $B S / S A$ can also be obtained using Ceva's theorem. Coordinate geometry solution can also be obtained.)
