Question

How to prove the basic proportionality theorem?

Answer 1

Answer:

given

• In trianhel in which de||bc
• trianhel in which de||bc construction : draw EN||AB
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CF
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×H
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×EN
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × EN
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × ENAB/EN EQ 1
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × ENAB/EN EQ 1AR TRIANGLE ADE1/2AE×DM
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × ENAB/EN EQ 1AR TRIANGLE ADE1/2AE×DMAR TRIANGLE 1/2 EC×DM
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × ENAB/EN EQ 1AR TRIANGLE ADE1/2AE×DMAR TRIANGLE 1/2 EC×DMAE/EC EQ2
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × ENAB/EN EQ 1AR TRIANGLE ADE1/2AE×DMAR TRIANGLE 1/2 EC×DMAE/EC EQ2AR TRIANGLE BDE=AR TRIANGLE CDE {TRIANGLES ON THE SAME BASE AND BETWEEN SAME PARALLEL ARE EQUAL IN AREA}
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × ENAB/EN EQ 1AR TRIANGLE ADE1/2AE×DMAR TRIANGLE 1/2 EC×DMAE/EC EQ2AR TRIANGLE BDE=AR TRIANGLE CDE {TRIANGLES ON THE SAME BASE AND BETWEEN SAME PARALLEL ARE EQUAL IN AREA}AD/DB=AR_EC ( BY EQ 1,2AND3 )
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × ENAB/EN EQ 1AR TRIANGLE ADE1/2AE×DMAR TRIANGLE 1/2 EC×DMAE/EC EQ2AR TRIANGLE BDE=AR TRIANGLE CDE {TRIANGLES ON THE SAME BASE AND BETWEEN SAME PARALLEL ARE EQUAL IN AREA}AD/DB=AR_EC ( BY EQ 1,2AND3 )HENCE PROVE
• trianhel in which de||bc construction : draw EN||ABDRAWDM||AC JOIN BE AND CFAREA OF TRIANGEL 1/2 B×HAR TRIANGLE ADE=1/2 AD×ENAR TRIANGLE BDE 1/2 DB × ENAB/EN EQ 1AR TRIANGLE ADE1/2AE×DMAR TRIANGLE 1/2 EC×DMAE/EC EQ2AR TRIANGLE BDE=AR TRIANGLE CDE {TRIANGLES ON THE SAME BASE AND BETWEEN SAME PARALLEL ARE EQUAL IN AREA}AD/DB=AR_EC ( BY EQ 1,2AND3 )HENCE PROVE

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.