<s>[INST] In square $ABCD$ with side $1$, points $E$, $P$, $F$ are the midpoints of $AD$, $CE$, $BP$. What is the area of $\triangle BFD$? I'm not sure if my solution to this Olympiad Geometry question is valid. 

Let $\square ABCD$ be a square with side length $1$. Let $E$, $P$, $F$ be the midpoints of $AD$, $CE$, $BP$, respectively. What is the area of $\triangle BFD$?

To solve this I used coordinate geometry but I'm not sure if it works.
I dropped a perpendicular from point P to side CD and called the point X. Then, since the area of triangle DPB is exactly double the desired area, I assumed point D to be (0, 0) and found the side length XP using the Pythagorean theorem. Since I have the three coordinates for triangle DPB, I used shoelace formula to find the area and divided by 2 to get {(19)^(1/2) - 1} / 8
 [/INST] The equation of the line $\overleftrightarrow{BD}$ is $x+y-1=0$. The coordinates of the point $F$ are $\left(\dfrac 34, \dfrac 38 \right)$. So the distance from the point $F$ to the line $\overleftrightarrow{BD}$ is 
$$h = \dfrac{\left|\dfrac 34 + \dfrac 38 - 1 \right|}{\sqrt{1^2+1^2}} 
  = \dfrac{1}{8 \sqrt 2}$$
The area of $\triangle BDF$ is therefore
$$\dfrac 12 \cdot BD \cdot h \ 
  = \dfrac 12 \cdot \sqrt 2 \cdot \dfrac{1}{8 \sqrt 2}
  = \dfrac{1}{16}$$
An easier-to-remember way to write @Rosenberg's answer is
$$BDF = \dfrac 12 \cdot
\left\| \begin{array}{c}
   1 & 1 & 1 \\
   1 & 0 & \dfrac 34 \\
   0 & 1 & \dfrac 38
\end{array} \right\| = \dfrac{1}{16}$$
where $\| \cdot \|$ indicates the absolute value of the determinant.

Notes. 
The distance from the point $(h,k)$ to the line $ax+by-c=0$ is
$\dfrac{|ah+bk-c|}{\sqrt{a^2+b^2}}$
The area of the triangle whose vertices have coordinates 
$(a_1, a_2), (b_1, b_2),(c_1, c_2)$ is
$$\dfrac 12 \cdot
\left\| \begin{array}{c}
   1   & 1   & 1 \\
   a_1 & b_1 & c_1 \\
   a_2 & b_2 & c_2
\end{array} \right\|$$
</s>