Question

proof of Basic Proportionality theorem​

Answer 1

Answer:

Basic proportionality theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. and DE intersects AB and AC at D and E respectively.

Answer 2

Hello,Buddy!!

Statement:-

• If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are are divided into equal ratio.

Given:-

• In ∆BC, DE||BC which intersect sides AB and AC at D & E respectively.

To Prove:-

• AD/DB = AE/EC.

Construction:-

• Draw DM⊥AC & EN⊥AB and join BE & CD.

Required Proof:-

WKT

• Area of Triangle = ½×(Base)×(Height)

Area of ∆ADE = ½×AD×EN

Area of ∆BDE = ½×BD×EN

→ ar(∆ADE)/ar(∆BDE)

→ (½×AD×EN)/(½×BD×EN)

→ AD/BD ...(1)

Similarly, Area of ∆ADE = ½×AE×DM

Area of ∆CDE = ½×EC×DM

→ ar(∆ADE)/ar(∆CDE)

→ (½×AE×DM)/(½×EC×DM)

→ AE/EC ...(2)

Now, ∆BDE & ∆CDE are on the same base (DE) and between same parallels (BC||DE).

→ ar(∆BDE) = ar(∆CDE) ...(3)

From (1),(2) & (3)

• AD/DB = AE/EC.

Hence, Proved!!

@MrMonarque♡

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