"given: in ∆abc, segment de is parallel to segment ac .. prove: bd over ba equals be over bc. the two-column proof with missing statements and reasons proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally.. statement.
reason. 1. segment de is parallel to segment ac. 1. given. 2. line segment ba is a transversal that intersects two parallel lines.
2. conclusion from statement 1. 3.. 3.. 4. ∡bde ≅ ∡bac. 4. reflexive property of equality. 5. ∆abc ~ ∆dbe. 5. angle-angle (aa) similarity postulate. 6.. 6.. "

Using the proportionality of sides we can writing 

BE/BC = BD/BA = DE/AC 

- you know that BA is a transversal that intersects two parallel
 lines =>m<D = m<A and m<E = m<C

so now there are 

m<D = m<A
m<E = m<E               => AAA similaritie case so triangle BDE similar triangle
<A common angle                                              BAC

hope helped 

Basically in step 4, they're using the symmetric property and not the reflexive property...not sure why they made this mistake
So to get to step 4, we need to get ∡BAC ≅ ∡BDE for step 3 (this is just the flipped version of step 4)
So the answer for the first part of step 3 is ∡BAC ≅ ∡BDE
The justification for this step is that a transversal line creates corresponding congruent angles

In step 5, we prove that the triangles are similar
So in step 6, we can say that the corresponding sides are proportional
Therefore, the answer for the first part in step 6 is
BD over BA equals BE over BC
and the justification for this step is

Side-Side-Side Similarity Theorem