prove that BPT or thales theorem
Answers
Answer:
Thales Theorem Statement
Let us now state the Basic Proportionality Theorem which is as follows:
If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Step-by-step explanation:
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.
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