What is bpt theoremwith proof
Answers
Basic Proportionality Theorem (can be abbreviated as BPT) states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
PROOF OF BPT
Given: In ΔABC, DE is parallel to BC
Line DE intersects sides AB and PQ in points D and E, such that we get triangles A-D-E and A-E-C.
To Prove: ADBD=AECE
Construction: Join segments DC and BE
Proof:
In ΔADE and ΔBDE,
A(ΔADE)A(ΔBDE)=ADBD (triangles with equal heights)
In ΔADE and ΔCDE,
A(ΔADE)A(ΔCDE)=AECE (triangles with equal heights)
Since ΔBDE and ΔCDE have a common base DE and have the same height we can say that,
A(ΔBDE)=A(ΔCDE)
Therefore,
A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)
Therefore,
ADBD=AECE
Hence Proved.
The BPT also has a converse which states, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
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.
