Explanation about BPT theorem
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State BPT :
If a line is drawn parallel to one side of a triangle to meet the other two sides in a distinct point,then the other two sides are divided into same ratio
Given: triABC in which DE||BC
to prove that: AD\DB= AE/EC
Proof:-
Construction- draw DM_|_ AC ,EN_|_ AB
Join BE &CD
Ar(ADE) = 1/2*AD*EN
Ar(BDE) = 1/2*DB*EN
Ar(ADE) div by ar(BDE) = 1/2AD.EN div by 1/2DB.EN = AD/DB ----------- eqn 1
Ar(ADE)=1/2*AE*DM
Ar(BEC)=1/2*EC*DM
Ar(ADE) div by ar(BEC) = 1/2AE.DM div by 1/2EC.DM = AE/EC (canceling 1/2 and DM)----------- eqn 2
Ar(BDE) = Ar(BEC) -------- eqn 3
(Triangle on the same base and between same parallel)
Therefore , AD/DB=AE\EC
If a line is drawn parallel to one side of a triangle to meet the other two sides in a distinct point,then the other two sides are divided into same ratio
Given: triABC in which DE||BC
to prove that: AD\DB= AE/EC
Proof:-
Construction- draw DM_|_ AC ,EN_|_ AB
Join BE &CD
Ar(ADE) = 1/2*AD*EN
Ar(BDE) = 1/2*DB*EN
Ar(ADE) div by ar(BDE) = 1/2AD.EN div by 1/2DB.EN = AD/DB ----------- eqn 1
Ar(ADE)=1/2*AE*DM
Ar(BEC)=1/2*EC*DM
Ar(ADE) div by ar(BEC) = 1/2AE.DM div by 1/2EC.DM = AE/EC (canceling 1/2 and DM)----------- eqn 2
Ar(BDE) = Ar(BEC) -------- eqn 3
(Triangle on the same base and between same parallel)
Therefore , AD/DB=AE\EC
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