Square geometry problem

We place the square in the coordinate plane as shown below:

The slope of segment EC is $\dfrac{4}{2.5} = \dfrac{8}{5}$ ; thus EC lies along the line $y - 4 = \dfrac{8}{5} (x - 4)$ , which can be re-written as $8x - 5y - 12 = 0$ .

The distance $D$ from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by

$D = \dfrac{\Big\lvert Ax_0 + By_0 + C \Big\rvert}{\sqrt{A^2 + B^2}}$

Thus in our case the distance from B $(0,4)$ to $8x - 5y - 12 = 0$ is

$\begin{aligned} d &= \dfrac{\Big\lvert 8(0) - 5(4) -12 \Big\rvert}{\sqrt{8^2 + (\text{-}5)^2}} \\ \\ &= \dfrac{32}{\sqrt{89}} \approx \boxed{3.39} \end{aligned}$

Note: Figure is not in scale

Using Pythagorean Theorem with $\triangle CDE$ , we can see that $EC=\sqrt{4^2+2.5^2}=\sqrt{22.25}$

Consider the area of $\triangle BEC$ $Area=\frac{1}{2}(4)(4)=8$

But at the same time, we can find the area with $Area=\frac{1}{2}(EC)(d)=8$

Therefore, the perpendicular distance from $B$ to line $EC$ is $d=\frac{16}{EC}=\frac{16}{\sqrt{22.25}}\approx\boxed{3.39}$

By pythagorean theorem, we have

$EC=\sqrt{(ED)^2+(CD)^2}=\sqrt{2.5^2+4^2}=\sqrt{22.25}$

Since $\angle DEC = \angle BCE$ , we have

$\dfrac{BF}{BC}=\dfrac{CD}{EC}$

$\dfrac{BF}{4}=\dfrac{4}{\sqrt{22.25}}$

$BF=\dfrac{4(4)}{\sqrt{22.25}} \approx \boxed{3.392}$

My solution via analytic geometry: At first, let's put our square into cartesian coordinate system, so that point A lies at the origin. Line EC is defined: $p:E\left[ {0;1.5} \right],C\left[ {4;4} \right]$ and has a corresponding vector: $p:\overrightarrow {EC} \in p,\overrightarrow {EC} = \left( {4 - 0;4 - 1.5} \right) = \left( {4;2.5} \right)$ General formula of a line is defined: $p:ax + by + c = 0$ Where a and b are perpendicular to the directional vector: $p:\overrightarrow n \bot \overrightarrow {EC} = \left( {2.5; - 4} \right)$ By plugging one of points lying on the line, we get c : $\begin{array}{l} p:E \in p,E\left[ {0;1.5} \right]\\ p:2.5x + \left( { - 4} \right)y + c = 0\\ p:2.5 \times 0 + \left( { - 4} \right) \times 1.5 + c = 0\\ p:c = 6 \end{array}$ So the full general formula of a line looks like this: $p:2.5x + \left( { - 4} \right)y + 6 = 0$ While having a general formula of a line and coordinates to the point, for which we want to know the distance to, we can use a distance formula: $v\left( {B,p} \right) = \frac{{\left| {a \times {b_1} + b \times {b_2} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$ Where: $B\left[ {{b_1};{b_2}} \right],\;p:ax + by + c = 0$ So the solution is: $v\left( {B,p} \right) = \frac{{\left| {2.5 \times 4 + \left( { - 4} \right) \times 0 + 6} \right|}}{{\sqrt {{{2.5}^2} + {{\left( { - 4} \right)}^2}} }}\sim\frac{{16}}{{4.716...}}\sim3.39cm$ And we can also check our solution graphically: