state and prove the basic Proportionality theorem and it's converse
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Answer:
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Step-by-step explanation:
Converse of Basic Proportionality Theorem. Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. ... Then there must be another line that is parallel to BC.
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Answer:
a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
If AD AE
---- = ------ then DE || BC
DB EC
Given : A Δ ABC and a line intersecting AB in D and AC in E,
such that AD / DB = AE / EC.
Prove that : DE || BC
Let DE is not parallel to BC. Then there must be another line that is parallel to BC.
Let DF || BC.
1) DF || BC (By assumption)
2) AD / DB = AF / FC (By Basic Proportionality
theorem)
3) AD / DB = AE /EC. (Given)
4) AF / FC = AE / EC. (By transitivity (from 2 and 3)
5) (AF/FC) + 1 = (AE/EC) + 1 (Adding 1 to both side)
6) (AF + FC )/FC = (AE + EC)/EC (By simplifying)
7) AC /FC = AC / EC (AC = AF +FC and AC= AE+EC)
8) FC = EC (As the numerator are same
so denominators are equal)
This is possible when F and E are same. So DF is the line DE itself.
∴ DF || BC
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