Basic proportionality theorem and it's converse
Answers
BASIC PROPORTIONALITY THEOREM (B.P.T)
THEOREM=IF A LINE PARALLEL TO A SIDE OF A TRIANGLE INTERSECTS THE REMAINING SIDES IN TWO DISTINCT POINTS, THEN THE LINE DIVIDES THE SIDES IN THE SAME PROPORTION.
GIVEN=in Δ ABC line l║line BC and line l intersects AB and AC in point P and Q respectively.
TO PROVE =AP/PB=AQ/QC
CONSTRUCTION=draw seg PC and seg QB
PROOF= Δ APQ and ΔPQB have equal heights.
A(ΔAPQ)/A(ΔPQB)=AP/PB ...(1) (areas proportionate to
bases)
and A(ΔAPQ)/A(ΔPQB)=AQ/QC ...(2) (areas proportionate to
bases)
∴ seg PQ is common base of ΔPQB and ΔPQC. Seg PQ ║seg BC. Hence Δ PQB andΔPQC have equal heights.
∴ A(ΔPQB) = A(ΔPQC) ...(3)
∴ A(ΔAPQ)/A(ΔPQB)=A(ΔAPQ)/A(ΔPQC) ...(From 1, 2 and 3)
∴AP/PB=AQ/QC ...(from 1 and 2)
CONVERSE OF BASIC PROPORTIONALITY THEOREM (B.P.T.)
THEOREM= IF THE LINE DIVIDES THE SIDES IN THE SAME PROPORTION THEN THE LINE PARALLEL TO A SIDE OF A TRIANGLE INTERSECTS THE REMAINING SIDES IN TWO DISTINCT POINTS.
(NOTE;CONVERSE OF B.P.T IS TOTALLY OPPOSITE OF THE ORIGINAL BASIC PROPORTIONALITY THEOREM . YOU CAN CHECK THE THEOREM SENTENCE FOR THE BETTER UNDERSTANDING)
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.
and
Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
If
AD AE
---- = ------ then DE || BC
DB EC
Given : A Δ ABC and a line intersecting AB in D and AC in E,
such that AD / DB = AE / EC.
Prove that : DE || BC
Let DE is not parallel to BC. Then there must be another line that is parallel to BC.
Let DF || BC.1) DF || BC 1) By assumption
2) AD / DB = AF / FC 2) By Basic Proportionality theorem
3) AD / DB = AE /EC 3) Given
4) AF / FC = AE / EC 4) By transitivity (from 2 and 3)
5) (AF/FC) + 1 = (AE/EC) + 1 5) Adding 1 to both side
6) (AF + FC )/FC = (AE + EC)/EC 6) By simplifying
7) AC /FC = AC / EC 7) AC = AF + FC and AC = AE + EC
8) FC = EC 8) As the numerator are same so denominators are equal
This is possible when F and E are same. So DF is the line DE itself.
∴ DF || BC
i hope it helps you.
