I'm just writing this problem out of curiosity, so let me know whether I make any mistakes. FYI this problem was inspired by the 1998 AIME problem 12. Also, if you have a better name for this question, please inform me.
Let $\Delta ABC$ be an equilateral triangle. Let $E$ be the midpoint of $\overline{AC}$, let $F$ be the midpoint of $\overline{AB}$, and let $D$ be the midpoint of $\overline{BC}$. Define a point $Q$, which lies on $\overleftrightarrow{EF}$ such that it does not coincide with either $E$ or $F$. Let $P$ be the intersection of the line through $E$ parallel to $\overline{BF}$ and line $\overline{BQ}$. Let $R$ be the intersection of lines $\overline{PA}$ and $\overline{CQ}$. Prove that:
1) \begin{equation} \frac{PQ}{QB} = \frac{PA}{AR} = \frac{CQ}{QR} = \frac{EP}{DE} = \frac{EQ}{QF} = \frac{FD}{FR} \end{equation}
- The area of $\Delta PQR$ is constant as $Q$ varies along $\overline{EF}$.
- What is the area of $\Delta PQR$?
I would love to see some type of synthetic solution.
This seemingly difficult problem can be solved by considering a pair of similarly placed similar triangles and the intercept theorem (Euclid VI. 2). We also need a lemma given below, proof of which is left to OP to work out.
$\underline{\text{Lemma}}:$
$\mathrm{Fig.\space 1}\space$ shows two triangles $ZXY$ and $VXW$, which share the vertex $X$. The sides opposite of vertex $X$ are parallel, i.e., $YZ \parallel VW$. Point $L$ lies on the side $YZ$ of $\triangle ZXY$. When the line $LX$ is extended, it intersects the side $VW$ of $\triangle VXW$ at $K$. It can be shown that, $$\dfrac{KW}{KV}=\dfrac{LY}{LZ}.$$
For ease of illustrating our solution, we divide it into three parts. For brevity, we denote the sidelength of the equilateral triangle $ABC$ as $2a$ and let $EQ=x$.
$\underline{\text{Part 1}}:$
Please pay your attention to $\mathrm{Fig.\space 2}$, in which we have marked point $G$ as the midpoint of the side $BC$. Out aim is to prove that (i) $CP$ is parallel to $AQ$ and (ii) the extended segment $GF$ passes through $R$.
Consider $\triangle BGP$. The line $FE$, which is parallel to its side $BG$, is intercepting its other two sides at $Q$ and $E$. According to the intercept theorem, we deduce, $$\dfrac{PG}{PE}=\dfrac{a}{x}\quad\rightarrow\quad \dfrac{PG}{PG-a}=\dfrac{a}{x}\quad\rightarrow\quad PG=\dfrac{a^2}{a-x} \quad\rightarrow\quad \dfrac{PG}{GC}=\dfrac{a}{a-x}. \tag{1}$$
Consider the two triangles $GCP$ and $FQA$, where we have $\measuredangle PGC=\measuredangle AFQ=60^o$. Furthermore, according to (1). They have sides about those equal angles proportional, i.e., $$\dfrac{PG}{GC}=\dfrac{AF}{FQ}=\dfrac{a}{a-x}.$$
According to the theorem Euclid VI. 6, triangles $GCP$ and $FQA$ are a pair of similar triangles. Besides, since $PG \parallel AF$ and $GC \parallel FQ$, they are similarly placed. Therefore, we can state that, not only $CP$ is parallel to $AQ$, but also the three lines joining their corresponding vertices i.e., $PA$, $CQ$, and $GF$, meet at $R$.
$\underline{\text{Part 2}}:$
Our attempt here is to prove that $AQ$ is parallel to $BR$. As shown in $\mathrm{Fig.\space 3}$, we have extended line $AQ$ to meet $FG$ at $S$. Now, we apply the lemma mentioned above to the two triangles $AQC$ and $SQR$, which share the vertex $Q$. We get, $$\dfrac{FS}{FR}=\dfrac{EA}{EC}=\dfrac{a}{a}=1\quad\rightarrow\quad FS=FR.\tag{2}$$ Because of (2) and $F$ is the midpoint of $AB$, quadrilateral $ARBQ$ is a parallelogram. This means $AQ \parallel BR$.
$\underline{\text{Part 3}}:$
As shown in $\mathrm{Fig.\space 4}$, we have added the line $AG$ to the diagram. Since $AQ \parallel BR$, we have $\triangle ARQ = \triangle ABQ$. Therefore, we can state, $$\triangle PRQ = \triangle AQP + \triangle ARQ = \triangle AQP + \triangle ABQ = \triangle ABP \tag{3}$$
Because of $PG \parallel AB$, we shall write, $$\triangle ABP = \triangle ABG = \dfrac{\triangle ABC}{2}. \tag{4}$$
Finally, from (3) and (4), it follows, $$\triangle PRQ = \dfrac{\triangle ABC}{2}. $$