$ABCD$ is a quadrilateral with right angles at $A$ and $C$. Points $E$ and $F$ are on $AC$, and $DE$ and $BF$ are perpendicular to $AC$. If $AE=3$, $DE=5$, and $CE=7$, then $BF=$

[asy]
draw((0,0)--(10,0)--(3,-5)--(0,0)--(6.5,3)--(10,0));
draw((6.5,0)--(6.5,3));
draw((3,0)--(3,-5));
dot((0,0));
dot((10,0));
dot((3,0));
dot((3,-5));
dot((6.5,0));
dot((6.5,3));
label("A", (0,0), W);
label("B", (6.5,3), N);
label("C", (10,0), E);
label("D", (3,-5), S);
label("E", (3,0), N);
label("F", (6.5,0), S);[/asy]

$\text{(A)} \ 3.6 \qquad \text{(B)} \ 4 \qquad \text{(C)} \ 4.2 \qquad \text{(D)} \ 4.5 \qquad \text{(E)} \ 5$
1. Given that $ABCD$ is a quadrilateral with right angles at $A$ and $C$, we can infer that $AC$ is a straight line and $DE$ and $BF$ are perpendicular to $AC$.
2. We are given the lengths $AE = 3$, $DE = 5$, and $CE = 7$. Since $E$ lies on $AC$, we have $AC = AE + EC = 3 + 7 = 10$.
3. We need to find the length of $BF$.

First, consider the triangles $\triangle EDC$ and $\triangle FCB$:
- $\triangle EDC$ has right angle at $E$ and $\triangle FCB$ has right angle at $F$.
- Since $DE$ and $BF$ are both perpendicular to $AC$, $\triangle EDC \sim \triangle FCB$ by AA similarity (both have a right angle and share $\angle DCE = \angle BCF$).

Using the similarity ratio:
\[
\frac{DE}{EC} = \frac{BF}{FC}
\]
Substituting the known values:
\[
\frac{5}{7} = \frac{BF}{FC}
\]

Next, consider the triangles $\triangle AFB$ and $\triangle DFA$:
- $\triangle AFB$ has right angle at $F$ and $\triangle DFA$ has right angle at $D$.
- Since $DE$ and $BF$ are both perpendicular to $AC$, $\triangle AFB \sim \triangle DFA$ by AA similarity (both have a right angle and share $\angle BAF = \angle DAE$).

Using the similarity ratio:
\[
\frac{AF}{BF} = \frac{AE}{DE}
\]
Substituting the known values:
\[
\frac{AF}{BF} = \frac{3}{5}
\]

Now, we have two equations:
1. $\frac{5}{7} = \frac{BF}{FC}$
2. $\frac{AF}{BF} = \frac{3}{5}$

From the first equation, solve for $FC$:
\[
FC = \frac{7}{5} BF
\]

From the second equation, solve for $AF$:
\[
AF = \frac{3}{5} BF
\]

Since $AC = AF + FC = 10$:
\[
AF + FC = 10
\]
Substitute the expressions for $AF$ and $FC$:
\[
\frac{3}{5} BF + \frac{7}{5} BF = 10
\]
Combine the terms:
\[
\frac{10}{5} BF = 10
\]
Simplify:
\[
2 BF = 10
\]
Solve for $BF$:
\[
BF = 5
\]

The final answer is $\boxed{5}$