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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. Complete the proof by entering the correct statements and reasons.

Mathematics
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|dw:1347551396246:dw|
What are the blanks? If you don't mind me asking.

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Other answers:

4th row we need to fill
so lets see whats happening in the next row - 5th one
4th and 6th
yea in 5th row, we are using AA similarity. which means we need two congruent Angles ?
Okay
did u understand 3rd row ?
Not really, I'm not good with Postulates.
|dw:1347552555431:dw|
look at the pix, DE is parallel to AC, its given
Yes, I understand that.
now lets do some imagination
Alright
think, you slide the side AC up and put it over DE
|dw:1347552973501:dw|
Got it.
then, since both lines are parallel, if you shrink AC, A overlaps wid D and C overlaps with E
right ?
Yes
since two sets of corresponding angles are equal, both triangles, small and big ones, are similar
Yes
|dw:1347553320251:dw|
so 4th statement can we have like this : < BED \(\cong\)
Agreed
|dw:1347553430140:dw|
good lets think of last row
Okay
last row must contain the thing asked for proving
Prove: BD over BA equals BE over BC
yea so lets put the last statement as \( \frac{BD}{BA} = \frac{BE}{BC} \)
can you think of what will be the Reason/justification ?
Um to be honest I have no idea...........Like I said I'm horrible at justifications.
So the answer is BDBA=BEBC "Corresponding sides of similar triangles are proportional"?

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