Theorem Statement: Thales’s Theorem or Basic Proportionality Theorem states that 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.
Proof of Basic Proportionality Theorem
Suppose we have a triangle ABC, if we draw a line LM parallel to side BC then the theorem states that,
In triangle ABC, join the vertex B to M on line AC and join vertex C on line AB. Then, drop a perpendicular MX on line AB and perpendicular LY on AC. The below diagram shows the construction for the same.
Since, area of triangle=
Area of ALM=
Area of LBM=
Area of ALM=
Area of LMC=
Ratio of area of ALM and LBM:
Ratio of area of ALM and LMC:
According to the property of triangles, the triangles on the same base and between same parallel lines have equal areas.
Therefore, LBM and LMC have equal areas.—-(3)
From equations (1),(2), and (3) we can conclude:
Hence, the basic proportionality theorem is proved.
Solved Examples on Basic Proportionality Theorem
Example 1. In an ∆ABC, sides AB and AC are intersected by a line at D and E respectively, which is parallel to side BC. Prove that AD/AB = AE/AC.
Given: DE || BC. So, AD/DB = AE/EC
or By interchanging the ratios as => DB/AD = EC/AE
Now, add 1 on both sides=> (DB/AD) + 1 = (EC/AE) + 1
(DB + AD)/AD = (EC + AE)/AE
AB/AD = AC/ AE
If we interchange the ratios again, we get=> AD/AB = AE/AC
Example 2. In triangle ABC, where DE is a line drawn from the midpoint of AB and ends midpoint of AC at E. AD/DB = AE/EC and ∠ADE = ∠ACB. Then prove ABC is an isosceles triangle.
Given: AD/DB = AE/EC
By the converse of the basic proportionality theorem, we get => DE || BC
According to question => ∠ADE = ∠ACB
Hence,∠ABC = ∠ACB
The side opposite to equal angles is also equal to AB = AC
Hence, ABC is an isosceles triangle.
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