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explain BPT theorem in simple method
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Basic Proportionality Theorem (Thales theorem): If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio.
In ∆ABC , if DE || BC and intersects AB in D and AC in E then
AD AE
---- = ------
DB EC
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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.
