1. **State the problem:** We have triangle $\triangle ABC$ with sides $BC=17$, $CA=18$, $AB=19$. Points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively. Point $P$ inside the triangle satisfies $PD \perp BC$, $PE \perp CA$, and $PF \perp AB$. Given $BD + CE + AF = 27$, we want to find $BD + BF$. 2. **Understand the configuration:** Since $PD$, $PE$, and $PF$ are perpendicular distances from $P$ to the sides, points $D$, $E$, and $F$ are the foots of perpendiculars from $P$ to the sides. 3. **Express variables:** Let $BD = x$, so $DC = 17 - x$. Similarly, let $CE = y$, so $EA = 18 - y$, and $AF = z$, so $FB = 19 - z$. 4. **Given sum:** We have $BD + CE + AF = x + y + z = 27$. 5. **Look for relation involving $BD + BF$:** We want $BD + BF = x + (19 - z) = x + 19 - z$. 6. **Use lemma about pedal triangles:** The points $D$, $E$, $F$ form the pedal triangle of $P$. A noteworthy property is that for a pedal triangle in $\triangle ABC$, sums of certain segments along the sides relate to the triangle's semiperimeter $s$. 7. **Calculate the semiperimeter:** $$s = \frac{17 + 18 + 19}{2} = \frac{54}{2} = 27.$$ 8. **Key insight:** For any point $P$ inside $\triangle ABC$, the sum $BD + CE + AF$ equals the semiperimeter $s$ of the triangle. Given here $BD + CE + AF = 27$, which matches $s$, this confirms the point $P$ is the incenter. 9. **Because $P$ is incenter:** Then $D$, $E$, and $F$ are the points where the incircle touches sides $BC, CA, AB$ respectively. 10. **Lengths of tangent segments from vertices to incircle touchpoints:** For incircle tangents: - $BD = BF = s - b$ where $b$ is length of side $AC$, - $CE = CE = s - c$, - $AF = AF = s - a$ where $a,b,c$ are the sides opposite respective vertices. Label sides opposite $A,B,C$ as $a=BC=17$, $b=CA=18$, $c=AB=19$. So: - $BD = BF = s - b = 27 - 18 = 9$. 11. **Find $BD + BF$:** Since $BD = BF = 9$, $$BD + BF = 9 + 9 = 18.$$