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.

My Personal Notes