To prove that the internal bisector $AD$ of triangle $ABC$ satisfies $\frac{BD}{DC}=\frac{AB}{AC}$, one draws through $C$ a line parallel to $DA$ meeting $BA$ produced at E. Which BPT conclusion in triangle BCE completes the proof?

In triangle $BCE$ the segment $DA$ is parallel to the side $CE$, so BPT gives $\frac{BD}{DC}=\frac{BA}{AE}$. Next, angle $AEC$ equals angle $BAD$ (corresponding angles) and angle $ACE$ equals angle $DAC$ (alternate angles); since $AD$ bisects angle A these two are equal, so triangle $ACE$ is isosceles with $AE=AC$. Substituting $AE=AC$ yields $\frac{BD}{DC}=\frac{AB}{AC}$. The other options pair the wrong segments or invent equalities such as $EA=BC$ that the construction does not give.