state bpt or phyta
theorem state and prove
Answers
Answer:
Basic proportionality theorem:
if a line parallel to a side of a triangle the remaining sides in two distinct points , then the line divides the sides in the same proportion.
given : in ∆abc line l || line bc and line l intersects ab and ac in point p and q respectively.
to prove : AP / PB = AQ / QC
construction : draw seg PC and seg BQ
proof : ∆ apq and ∆ pqb have equal heights .
A( ∆ apq ) / A( ∆ pqb ) = AP / PB......(1) ( areas proportionate to base )
A( ∆ apq ) / A( ∆ pqc ) = AQ / QC.....(2) ( areas proportionate to base )
seg pq is common base of ∆ pqb and ∆ pqc. seg pq || seg bc , hence ∆ pqb and ∆ pqc have equal heights .
A( ∆ pqb ) = A( ∆ pqc ) .......(3)
A( ∆ apq ) / A( ∆ pqb ) = A( ∆ apq ) / A( ∆ pqc )........ from (1) (2) and (3)
therefore , AP / PB = AQ / QC ....from(1) & (2). hence proved.
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Answer:-
PROOF OF BPT
Given: In ΔABC, DE is parallel to BC
Line DE intersects sides AB and AC in points D and E respectively.
To Prove: => AD/DB = AE/AC
Construction: Draw EF ⟂ AD and DG⟂ AE and join the segments BE and CD.
Proof:
Area of Triangle= ½ × base × height
In ΔADE and ΔBDE,
=> Ar(ADE) / Ar(DBE)
= ½ ×AD×EF / ½ ×DB×EF
= AD/DB ......(1)
In ΔADE and ΔCDE,
=> Ar(ADE)/Ar(ECD)
= ½×AE×DG / ½×EC×DG
= AE/EC ........(2)
Note => that ΔDBE and ΔECD have a common base DE and lie between the same parallels DE and BC. Also, we know that triangles having the same base and lying between the same parallels are equal in area.
So, we can say that
Ar(ΔDBE)=Ar(ΔECD)
Therefore,
A(ΔADE)/A(ΔBDE) = A(ΔADE)/A(ΔCDE)
Therefore,
=> AD/DB = AE/AC
Hence Proved.
i hope it helps you.
