Let $ABC$ be a triangle with $AB = 3$, $BC = 4$, and $AC =5$. If $D$ is the projection from $B$ onto $AC$, $E$ is the projection from $D$ onto $BC$, and $F$ is the projection from $E$ onto $AC$, compute the length of the segment $DF$.

[i]2016 CCA Math Bonanza Individual #5[/i]
1. **Determine the length of \(BD\):**
   - Since \(D\) is the projection of \(B\) onto \(AC\), \(BD\) is the altitude from \(B\) to \(AC\).
   - Using the formula for the altitude in a right triangle, we have:
     \[
     BD = \frac{AB \times BC}{AC} = \frac{3 \times 4}{5} = \frac{12}{5}
     \]

2. **Determine the length of \(DC\):**
   - Since \(D\) lies on \(AC\), we can use the Pythagorean theorem in \(\triangle BDC\):
     \[
     DC = \sqrt{BC^2 - BD^2} = \sqrt{4^2 - \left(\frac{12}{5}\right)^2} = \sqrt{16 - \frac{144}{25}} = \sqrt{\frac{400}{25} - \frac{144}{25}} = \sqrt{\frac{256}{25}} = \frac{16}{5}
     \]

3. **Determine the length of \(DE\):**
   - Since \(E\) is the projection of \(D\) onto \(BC\), \(DE\) is the altitude from \(D\) to \(BC\).
   - Using the formula for the altitude in a right triangle, we have:
     \[
     DE = \frac{BD \times DC}{BC} = \frac{\frac{12}{5} \times \frac{16}{5}}{4} = \frac{\frac{192}{25}}{4} = \frac{48}{25}
     \]

4. **Determine the length of \(EC\):**
   - Since \(E\) lies on \(BC\), we can use the Pythagorean theorem in \(\triangle DEC\):
     \[
     EC = \sqrt{DC^2 - DE^2} = \sqrt{\left(\frac{16}{5}\right)^2 - \left(\frac{48}{25}\right)^2} = \sqrt{\frac{256}{25} - \frac{2304}{625}} = \sqrt{\frac{2560}{625} - \frac{2304}{625}} = \sqrt{\frac{256}{625}} = \frac{16}{25}
     \]

5. **Determine the length of \(EF\):**
   - Since \(F\) is the projection of \(E\) onto \(AC\), \(EF\) is the altitude from \(E\) to \(AC\).
   - Using the formula for the altitude in a right triangle, we have:
     \[
     EF = \frac{EC \times DE}{DC} = \frac{\frac{16}{25} \times \frac{48}{25}}{\frac{16}{5}} = \frac{\frac{768}{625}}{\frac{16}{5}} = \frac{768}{625} \times \frac{5}{16} = \frac{768}{10000} = \frac{192}{125}
     \]

6. **Determine the length of \(DF\):**
   - Since \(F\) lies on \(AC\), we can use the Pythagorean theorem in \(\triangle DEF\):
     \[
     DF = \sqrt{DE^2 - EF^2} = \sqrt{\left(\frac{48}{25}\right)^2 - \left(\frac{192}{125}\right)^2} = \sqrt{\frac{2304}{625} - \frac{36864}{15625}} = \sqrt{\frac{36864}{15625} - \frac{36864}{15625}} = \sqrt{0} = 0
     \]

The final answer is \(\boxed{0}\)