Math, asked by kanharai2004p2yq3s, 1 year ago

how to prove basic proportionality theorem

Answers

Answered by badal968
4
it is a bpt theorem
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Answered by nilesh102
0

hi mate,

PROOF OF BPT

Given: In ΔABC, DE is parallel to BC

let PQ = DE

Line DE intersects sides AB and AC in points D and E respectively.

To Prove:

AD AE

----- = -----

DB 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) ½ ×AD×EF AD

----------- = ------------------ = ------ .....(1)

Ar(DBE) ½ ×DB×EF DB

In ΔADE and ΔCDE,

Ar(ADE) ½×AE×DG AE

------------ = --------------- = ------ ........(2)

Ar(ECD) ½×EC×DG EC

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(ΔADE)

------------- = ---------------

A(ΔBDE) A(ΔCDE)

Therefore,

AD AE

----- = -----

DB AC

Hence Proved.

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