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OpenStudy (alexandervonhumboldt2):
yes
OpenStudy (anonymous):
In ΔABC shown below, BD over BA equals BE over BC:
Triangle ABC with segment DE intersecting sides AB and BC respectively.
The flowchart proof with missing statements and reasons proves that if a line intersects two sides of a triangle and divides these sides proportionally, the line is parallel to the third side:
OpenStudy (anonymous):
@Abhisar
OpenStudy (anonymous):
@jim_thompson5910 Please Help!
OpenStudy (anonymous):
@ganeshie8
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OpenStudy (anonymous):
Anyone?
OpenStudy (anonymous):
@kmorgan
OpenStudy (anonymous):
She is offline, why would you tag her lol
OpenStudy (anonymous):
idk
OpenStudy (anonymous):
@Ashleyisakitty
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OpenStudy (anonymous):
@awkwardpanda @SolomonZelman @zepdrix
jimthompson5910 (jim_thompson5910):
notice how the second block has "side angle side similarity postulate"
jimthompson5910 (jim_thompson5910):
the proportion BD/BA = BE/BC takes care of the two "S"s for the sides
you just need a statement and reason that helps with the angle portion
OpenStudy (anonymous):
1. ∠A ≅ ∠A
2. Reflexive Property of Equality
1. ∠A ≅ ∠B
2. Corresponding Parts of Similar Triangles
1. ∠A ≅ ∠B
2. Corresponding Angles Postulate
1. ∠B ≅ ∠B
2. Reflexive Property of Equality
jimthompson5910 (jim_thompson5910):
|dw:1419128066396:dw|
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