Math, asked by punya30a, 1 year ago

proof of bpt theoram​

Answers

Answered by DSaiKiran
7

Step-by-step explanation:

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.

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.

Attachments:
Answered by meghaundirwade
2

Answer:

in BPT theorem If a line parallel to side of a triangle intersects the remaining sides in two distinct points then the line divide the side in the same proportion

given in triangle abc line l parallel to line BC and line l intersect ab and ac at point P and Q respectively

to prove AP by PB is equal to 1 by QC

construction draw vi B Q and Pc

PROOF :in triangle A P Q and triangle pbq

area of triangle A P Q by area of triangle pqr is equal to AP by PB reason ratio of area of two Triangles have common height _(1)

area of triangle A P Q by area of triangle pqr is equal to x cube IQ reason ratio of area of two triangle have common height _(2)

PQ is a common base of triangle pqr and triangle pqc, seg PQ parallel line BC

Hans triangle pqr and triangle pqc is equal height

area of triangle pbq is equal to area of triangle pqc _(3)

from equation 1, 2 and 3 we get,

area of triangle A P Q by area of triangle pqr is equal to area of triangle A P Q by area of triangle pqc

AP by PB is equal to 1 cubic QC.

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