In square $ABCD$ with side $1$, points $E$, $P$, $F$ are the midpoints of $AD$, $CE$, $BP$. What is the area of $\triangle BFD$?

Question

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

Answer 1

Let $A(0,0)$, $B(0,1),$ $C(1,1)$ and $D(1.0).$

Thus, $E(0.5,0)$, $P(0.75,05)$ and $F(0.375,0.75).$

Now, use $$S_{\Delta BFD}=\frac{1}{2}\left|x_B(y_F-y_D)+x_F(y_D-y_B)+x_D(y_B-y_F)\right|.$$ I got $$S_{\Delta BDF}=\frac{1}{16}.$$

Answer 2

Using Euclidean geometry you get straight to the result.

Distance of $P$ from $DC$ is $\frac{1}{4}$ and distance of $P$ from $BC$ is $\frac{1}{2}$, by Thales theorem. So you obtain the area of $DPB$ by successive subtractions from the area of $ABCD$, then divide by two to get the desired area: \begin{eqnarray} \mathcal A_{BFD}&=& \frac{1}{2}\cdot\left(\mathcal A_{ABCD}-\mathcal A_{ABD}- \mathcal A_{BCP}-\mathcal A_{CDP}\right)=\\ &=& \frac{1}{2}\cdot\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}\right)=\\ &=&\frac{1}{16}. \end{eqnarray}

$\hskip1.5in$

Answer 3

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 $$\dfrac 12 \cdot \left\| \begin{array}{c} 1 & 1 & 1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array} \right\|$$