Rectangle $ABCD$ has $AB=4$ and $BC=3$. Segment $EF$ is constructed through $B$ so that $EF$ is perpendicular to $DB$, and $A$ and $C$ lie on $DE$ and $DF$, respectively. What is $EF$?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ \frac {125}{12} \qquad \mathrm{(D)}\ \frac {103}{9} \qquad \mathrm{(E)}\ 12$
Solution 1
The situation is shown in the picture below.

From the [Pythagorean theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_theorem) we have $BD=5$.
Triangle $EAB$ is similar to $BAD$, as they have the same angles. Segment $BA$ is perpendicular to $DA$, meaning that angle $DAB$ and $BAE$ are right angles and congruent. Also, angle $DBE$ is a right angle. Because it is a rectangle, angle $BDC$ is congruent to $DBA$ and angle $ADC$ is also a right angle. By the transitive property:
$mADB + mBDC = mDBA + mABE$
$mBDC = mDBA$
$mADB + mBDC = mBDC + mABE$
$mADB = mABE$
Next, because every triangle has a degree measure of $180$, angle $BEA$ and angle $DBA$ are similar. 

Hence $BE/AB = DB/AD$, and therefore $BE = AB\cdot DB/AD = 20/3$.
Also triangle $CBF$ is similar to $ABD$. Hence $BF/BC = DB/AB$, and therefore $BF=BC\cdot DB / AB = 15/4$.
We then have $EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}$.

Solution 2
Since $BD$ is the altitude from $D$ to $EF$, we can use the equation $BD^2 = EB\cdot BF$. 
Looking at the angles, we see that triangle $BDE$ is similar to $DCB$. Because of this, $\frac{AB}{CB} = \frac{EB}{DB}$. From the given information and the [Pythagorean theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_theorem), $AB=4$, $CB=3$, and $DB=5$. Solving gives $EB=20/3$. 
We can use the above formula to solve for $BF$. $BD^2 = 20/3\cdot BF$. Solve to obtain $BF=15/4$. 
We now know $EB$ and $BF$. $EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}$.

Solution 3(Coordinate Bash)
To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the $x$-axis.It is also worth noting the $F$ will lie on the $x$ axis and $E$ on the $y$. Let $D$ be the origin, $A(3,0)$, $C(4,0)$, and  $B(4,3)$. We can express segment $DB$ as the line $y=\frac{3x}{4}$. 
Since $EF$ is perpendicular to $DB$, and we know that $(4,3)$ lies on it, we can use this information to find that segment $EF$
is on the line $y=\frac{-4x}{3}+\frac{25}{3}$. Since $E$ and $F$ are on the $y$ and $x$ axis, respectively, we plug in $0$ for $x$
and $y$, we find that point $E$ is at $(0,\frac{25}{3})$, and point $F$ is at $(\frac{25}{4},0)$. Applying the distance formula,
we obtain that $EF$= $\boxed{\frac{125}{12}}$.