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.

- anonymous

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- anonymous

##### 2 Attachments

- anonymous

|dw:1347551396246:dw|

- anonymous

What are the blanks? If you don't mind me asking.

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## More answers

- ganeshie8

4th row we need to fill

- ganeshie8

so lets see whats happening in the next row - 5th one

- anonymous

4th and 6th

- ganeshie8

yea in 5th row, we are using AA similarity.
which means we need two congruent Angles ?

- anonymous

Okay

- ganeshie8

did u understand 3rd row ?

- anonymous

Not really, I'm not good with Postulates.

- ganeshie8

|dw:1347552555431:dw|

- ganeshie8

look at the pix,
DE is parallel to AC, its given

- anonymous

Yes, I understand that.

- ganeshie8

now lets do some imagination

- anonymous

Alright

- ganeshie8

think, you slide the side AC up and put it over DE

- ganeshie8

|dw:1347552973501:dw|

- anonymous

Got it.

- ganeshie8

then, since both lines are parallel,
if you shrink AC,
A overlaps wid D and C overlaps with E

- ganeshie8

- ganeshie8

right ?

- anonymous

Yes

- ganeshie8

since two sets of corresponding angles are equal,
both triangles, small and big ones, are similar

- anonymous

Yes

- ganeshie8

|dw:1347553320251:dw|

- ganeshie8

so 4th statement can we have like this :
< BED \(\cong\)

- anonymous

Agreed

- ganeshie8

|dw:1347553430140:dw|

- ganeshie8

good lets think of last row

- anonymous

Okay

- ganeshie8

last row must contain the thing asked for proving

- anonymous

Prove: BD over BA equals BE over BC

- ganeshie8

yea so lets put the last statement as
\( \frac{BD}{BA} = \frac{BE}{BC} \)

- ganeshie8

can you think of what will be the Reason/justification ?

- anonymous

Um to be honest I have no idea...........Like I said I'm horrible at justifications.

- anonymous

So the answer is BDBA=BEBC "Corresponding sides of similar triangles are proportional"?

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