Question 153
Suppose point $P$ is inside triangle $ABC$. Let $AP$, $BP$, and $CP$ intersect sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Suppose $\angle APB = \angle BPC = \angle CPA$, $PD = \frac{1}{4}$, $PE = \frac{1}{5}$, and $PF = \frac{1}{7}$. Compute $AP + BP + CP$.