Question

Prove the Basic Proportionality Theorem (BPT)

Answer 1

Answer:

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Answer 2

hi mate,

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 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.

i hope it helps you.