Question

frd pls tell me BPT theorem in maths
​

Answer 1

Answer:

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 AC in points D and E respectively.

To Prove: ADBD=AECE

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)=12×AD×EF12×DB×EF=ADDB(1)

In ΔADE and ΔCDE,

Ar(ADE)Ar(ECD)=12×AE×DG12×EC×DG=AEEC(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,

ADBD=AECE

Hence Proved.

PLZ MARK IT AS BRAINLIEST

Answer 2

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.