prove bpt rule
class 10
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Answer:
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Step-by-step explanation:
BPT theorem :- " If a line is drawn parallel to one side of a triangle interacting the other two sides in distinct points, then the other two sides are divided in the same ratio."
Q. GIVEN:- In the ∆ ABC's BC ll DE .
WHAT TO PROVE :- We have to prove that AD\DB = AE/EC
CONSTRUCTION :- Join D,C and B,E. Draw EG perpendicular AD and DF perpendicular AE.
prove:- Now,
We found in the ∆ADE -----
ar(ADE) = 1/2 × b × h
= 1/2 × AD × EG. ⟨--------- (I)
and ar(ADE) = 1/2 × b × h
=. 1/2 × AE × DF ⟨-------- (ii)
Again,
We found in the ∆BDE and ∆ DEC -----
ar(BDE) = 1/2 × b × h
= 1/2 × DB × EG ⟨---------- (iii)
and , ar(DEC) = 1/2 × b × h
= 1/2× EC × DF⟨---(iv)
Now,
Deviding (I) by (iii)-----
ar(ADE)/ar(BDE)= 1/2×AD×EG/ 1/2×DB×EG
= AD/DB ⟨----- (V)
again,
Deviding (ii) by (iv)-----
ar(ADE)/ar(DEC)= 1/2×AE×DF/ 1/2×EC×DF
= AE/EC ------(VI)
We know that , ∆BDE and ∆DEC's base is same and DE||BC.
Therefore,
ar(BDE)=ar(DEC) ---(vii)
From (v),(vi) and (vii) we get-----
AD/DB = AE/EC.
PROVED .
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