BIGDOG96
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.
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BIGDOG96
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TheMind
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BIGDOG96
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What are the blanks? If you don't mind me asking.
ganeshie8
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4th row we need to fill
ganeshie8
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so lets see whats happening in the next row - 5th one
BIGDOG96
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4th and 6th
ganeshie8
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yea in 5th row, we are using AA similarity.
which means we need two congruent Angles ?
BIGDOG96
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Okay
ganeshie8
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did u understand 3rd row ?
BIGDOG96
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Not really, I'm not good with Postulates.
ganeshie8
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ganeshie8
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look at the pix,
DE is parallel to AC, its given
BIGDOG96
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Yes, I understand that.
ganeshie8
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now lets do some imagination
BIGDOG96
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Alright
ganeshie8
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think, you slide the side AC up and put it over DE
ganeshie8
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BIGDOG96
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Got it.
ganeshie8
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then, since both lines are parallel,
if you shrink AC,
A overlaps wid D and C overlaps with E
ganeshie8
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<A = <D
ganeshie8
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<C = <E
ganeshie8
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right ?
BIGDOG96
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Yes
ganeshie8
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since two sets of corresponding angles are equal,
both triangles, small and big ones, are similar
BIGDOG96
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Yes
ganeshie8
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ganeshie8
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so 4th statement can we have like this :
< BED \(\cong\) <BCA by "Corresponding Angles Postulate"
?
BIGDOG96
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Agreed
ganeshie8
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ganeshie8
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good lets think of last row
BIGDOG96
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Okay
ganeshie8
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last row must contain the thing asked for proving
BIGDOG96
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Prove: BD over BA equals BE over BC
ganeshie8
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yea so lets put the last statement as
\( \frac{BD}{BA} = \frac{BE}{BC} \)
ganeshie8
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can you think of what will be the Reason/justification ?
BIGDOG96
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Um to be honest I have no idea...........Like I said I'm horrible at justifications.
BIGDOG96
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So the answer is BDBA=BEBC "Corresponding sides of similar triangles are proportional"?