Question

HEY CAN ANYONE HELP ME!



In ΔABC shown below, segment DE is parallel to segment AC:

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. Line segment DE is parallel to line segment AC 1. Given
2. Line segment AB is a transversal that intersects two parallel lines. 2. Conclusion from Statement 1.
3. 3.
4. ∠B ≅ ∠B 4. Reflexive Property of Equality
5. 5.
6. BD over BA

Answer 1

@demonchild99 where's the triangle???

Answer 2

oh hold up...my bad

Answer 3

ok

Answer 4

@iGreen

Answer 5

The table is kind of scrambled..can you fix it?

Answer 6

i can try

Answer 7

Nevermind I found this:

"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"

http://openstudy.com/study#/updates/4fd57661e4b04bec7f16ab5e

Answer 8

If you're seeing those question marks just click on a different post and click back here.

Answer 9

so wait...ok....would that mean the answer is sss similarity theorem?

Answer 10

For those who see this and was as confused as I was it is
3. ∠ BDE ≅ ∠ BAC; Corresponding Angles Postulate
5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate