The proof of BPT uses the step $\frac{ar(ADE)}{ar(BDE)}=\frac{AD}{DB}$. Why does this ratio of areas equal the ratio of the two segments?

Triangles ADE and BDE sit on the bases $AD$ and $DB$ of the same straight line $AB$, and both have their third vertex at E, so the perpendicular from E to $AB$ is a common height $h$. Their areas are $\frac{1}{2}\times AD\times h$ and $\frac{1}{2}\times DB\times h$, whose ratio is $\frac{AD}{DB}$. They are neither similar nor congruent in general, and they certainly do not have equal areas unless $AD=DB$ — the pair with equal areas in this proof is BDE and CDE, which lie on the common base DE between the parallels DE and BC.