Statement:
BPT (Basic Proportionality Theorem),
If a line is drawn parallel to one of the triangle to intersects the other two sides in distinct points,
then the other two sides of
the triangle are divided into the same ratio.
Proof:
Given:
In ∆ABC, DE || BC and AB and AC are intersected by DE at points D and E respectively.
To prove:
AD / DB = AE / EC
Construction:
Join BE and CD.and
Draw:
EG⊥AB and DF⊥AC
Proof:
We know that
ar( Δ ADE) = 1 / 2 × AD × EG
ar( Δ DBE) = 1 / 2 × DB × EG
So
ar(Δ ADE) / ar(Δ DBE) =
AD / DB ------- (1)
Similarly,
ar(Δ ADE) / ar(Δ ECD) =
AE / EC ----------(2)
Now,
Δ DBE and Δ ECD
are the on the same base DE and also between the same parallels i.e. DE and BC,
So
ar(Δ DBE) = ar(Δ ECD) ---(3)
By (1), (2) , (3)
AD / DB = AE / EC
Hence proved.
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