Question

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Here's my question....


..

PROVE THE BASIC PROPORTIONALITY THEOREM????

CLASS 10..✌NO SPAM...​

Answer 1

Given :-

In △ ABC, DE || BC

Line DE intersect at sides AB & AC.

To proof :- AD/ BD = AE/CE

Construction :- Draw EF perpendicular to AD and DG perpendicular to AE, Join BE and CD.

Proof :-

Area of triangle = 1/2 × base × height

In △ADE and △ BDE,

ar(ADE)/ar(BDE) = 1/2 × AD × EF / 1/2 × DB × EF

= AD/DB ---(1)

In ΔADE and ΔCDE,

ar (ADE)/ ar (CDE) = 1/2 × AE × DG/ 1/2 × EC × DG

= AE/EC ---(2)

△DBE and △ECD have same base DE and lies between same parallel line DE and BC. Triangle on same base and between same parallel line have equal area.

Therefore,

ar ( DBE) = ar ( ECD)

Therefore,

ar (ADE)/ ar (BDE) = ar (ADE) / ar (CDE)

Therefore,

AD/BD = AE/CE

Hence proved

Refer attachment for diagram.

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: 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.