Question

state and prove BPT
​

Answer 1

basic proportionality

MATHS

State and prove BPT.

December 20, 2019avatar

Pragya Dwivedi

SHARE

ANSWER

Statement : If a line passing through two sides of triangle then it os parallel to third side then divides other two sides in same ratio.

In ΔADE (Baes AD)

Area of triangle =

2

1

.AD.ME _______ (1)

Again in ΔADE (Base AE)

ar.(ΔADE)=

2

1

.AE.ND ________ (2)

In ΔBDE (Base BD)

ar.(ΔBDE)=

2

1

.DB.ME ________ (3)

In ΔDEC (Base EC)

ar.(ΔDEC)=

2

1

.EC.ND ________ (4)

Equation (1) & (3)

ar.ΔBDE

ar.ΔADE

=

2

1

.DB.ME

2

1

.AD.ME

ar.ΔBDE

arΔADE

=

DB

AD

_______ (5)

Now equation (2) & (4)

ar.(ΔDEC)

ar.(ΔADE)

=

2

1

.EC.ND

2

1

.AE.ND

ΔDEC

ar.(ΔADE)

=

EC

AE

_______ (6)

By theorem,

ar.(ΔBDE)=ar.(ΔDEC)

ar.(ΔBDC)

ar.(ΔADE)

=

ar.(ΔDEC)

ar.(ΔADE)

=

BD

AD

ar.(ΔDEC)

ar.(ΔADE)

=

BD

AD

________ (7)

By equation (6) & (7)

There L.H.S is same and R.H.S is same.

EC

AE

=

BD

AD

∴

DB

AD

=

EC

AE

plzz mark as brainlist

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.