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Thales’s Theorem
  • Last Updated : 19 Jan, 2021

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,

\frac{AP}{PB}=\frac{AQ}{QC}    

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= (\frac{1}{2}*base*height)

Area of \triangle   ALM=(\frac{1}{2}*AL*MX)

Area of \triangle   LBM=(\frac{1}{2}*LB*MX)

Area of \triangle   ALM=(\frac{1}{2}*AM*LY)

Area of \triangle   LMC=(\frac{1}{2}*MC*LY)

Ratio of area of \triangle   ALM and \triangle   LBM:

\frac{area(\triangle ALM)}{area(\triangle LBM)}=\frac{(\frac{1}{2}*AL*MX)}{(\frac{1}{2}*LB*MX)}=\frac{AL}{LB}   —-(1)

Ratio of area of \triangle   ALM and \triangle   LMC:



\frac{area(\triangle ALM)}{area(\triangle LMC)}=\frac{(\frac{1}{2}*AM*LY)}{(\frac{1}{2}*MC*LY)}=\frac{AM}{MC}   —-(2)

According to the property of triangles, the triangles on the same base and between same parallel lines have equal areas.

Therefore, \triangle   LBM and \triangle   LMC have equal areas.—-(3)

From equations (1),(2), and (3) we can conclude:

\frac{AL}{LB}=\frac{AM}{MC}

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.

Solution:

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

Hence, proved.

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.

Solution:

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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