Question

Given: In ∆ABC, segment DE is parallel to segment AC .
Prove: BD over BA equals BE over BC

Triangles ABC and DBE where DE is parallel to AC.

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. ∡BDE ≅ ∡BAC
3. Corresponding Angles Postulate
4.
4.
5. ∆ABC ~ ∆DBE
5. Angle-Angle (AA) Similarity Postulate
6.
6.

Complete the proof by entering the correct statements and reasons.

Answer 1

so this isn't multiple choice and it's more like type in the answer?

Answer 2

An essay type question.

Answer 3

yeah just need to think of the right wording, one sec

Answer 4

Okay.

Answer 5

ugh, had a glitch, one sec while I retype, so sorry about that

Answer 6

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


again sry about that

Answer 7

Totally okay.(: