### 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:

—-(1)

Ratio of area of ALM and LMC:

—-(2)

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.**

**Solution:**

Given:DE || BC. So, AD/DB = AE/ECor 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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