In the figure, a point O inside ∆ABC is joined to its vertices. From a point D on AO, DE is drawn parallel to AB & from E, EF is drawn parallel to BC. Prove that DF || AC.
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BASIC PROPORTIONALITY THEOREM (BPT) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. This is also known as Thales 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.
SOLUTION:
In ΔOAB, DE || AB (Given)
OD/DA= OE/EB ………..(i)
[By Basic Proportionality Theorem]
In ΔOBC,EF || BC (Given)
OE/EB= OF/FC…………(ii)
[By Basic Proportionality Theorem]
From eq (i) and (ii),
OD/DA = OF/FC
D & F are points on sides OA & OC of ∆OAC,such that
OD/DA = OF/FC
DF || AC
[By converse of Basic Proportionality Theorem].
Hence, proved
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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.
SOLUTION:
In ΔOAB, DE || AB (Given)
OD/DA= OE/EB ………..(i)
[By Basic Proportionality Theorem]
In ΔOBC,EF || BC (Given)
OE/EB= OF/FC…………(ii)
[By Basic Proportionality Theorem]
From eq (i) and (ii),
OD/DA = OF/FC
D & F are points on sides OA & OC of ∆OAC,such that
OD/DA = OF/FC
DF || AC
[By converse of Basic Proportionality Theorem].
Hence, proved
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