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Basic proportionality theorem or Thales theorem

**Theorem 1**: A straight line is drawn parallel to one side of a triangle intersecting the other two sides divides the sides in the same ratio.

**Given**: In a triangle, \(ABC\), \(D\) is a point on \(AB\) and \(E\) is a point on \(AC\).

**To prove**: \(\frac{AD}{DB} = \frac{AE}{EC}\)

**Construction**: Draw a line \(DE \parallel BC\).

**Proof**: Here, \(\angle ABC = \angle ADE = \angle 1\) [Corresponding angles are equal because \(DE \parallel BC\)]

\(\angle ACB = \angle AED = \angle 2\) [Corresponding angles are equal because \(DE \parallel BC\)]

\(\angle BAC = \angle DAE = \angle 3\) [Both triangles have a common angle]

\(\triangle ABC \sim \triangle ADE\) [By AAA similarity]

\(\frac{AB}{AD} = \frac{AC}{AE}\) [Corresponding sides are proportional]

\(\frac{AD + DB}{AD} = \frac{AE + EC}{AE}\)

\(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\)

\(\frac{DB}{AD} = \frac{EC}{AE}\)

\(\frac{AD}{DB} = \frac{AE}{EC}\) [Taking reciprocal]

Hence, we proved.

**Corollary**:

In a \(\triangle ABC\), a straight line \(DE\) intersects \(AB\) at \(D\) and \(AC\) at \(E\) and is parallel to \(BC\), then prove that

(i) \(\frac{AB}{AD} = \frac{AC}{AE}\) (ii) \(\frac{AB}{DB} = \frac{AC}{EC}\)

**Proof**:

(i) By Thales theorem, we have:

\(\frac{AD}{DB} = \frac{AE}{EC}\)

\(\frac{DB}{AD} = \frac{EC}{AE}\)

Adding \(1\) on both sides of the equation, we have:

\(\frac{DB}{AD} + 1 = \frac{EC}{AE} + 1\)

\(\frac{DB + AD}{AD} = \frac{EC + AE}{AE}\)

\(\frac{AB}{AD} = \frac{AC}{AE}\)

Hence, we proved.

(ii) By Thales theorem, we have:

\(\frac{AD}{DB} = \frac{AE}{EC}\)

Adding \(1\) to both sides of the equation, we have:

\(\frac{AD}{DB} + 1 = \frac{AE}{EC} + 1\)

\(\frac{AD + DB}{DB} = \frac{AE + EC}{EC}\)

\(\frac{AB}{DB} = \frac{AC}{EC}\)

Hence, we proved.

Converse of Basic proportionality theorem or Converse of Thales theorem.

**Theorem 2**: If a straight line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

**Given**: In \(\triangle ABC\), \(\frac{AD}{DB} = \frac{AE}{EC}\)

**To prove**: \(DE \parallel BC\)

**Construction**: If \(DE\) is not parallel to \(BC\), then draw \(BF \parallel DE\).

**Proof**: Since \(DE \parallel BC\), and two sides \(AB\) and \(AC\) are divided in the same ratio, then by Thales theorem, we get:

\(\frac{AD}{DB} = \frac{AE}{EC}\) ---- (\(1\))

\(\frac{AD}{DB} = \frac{AF}{FC}\) ---- (\(2\))

From equations (\(1\)) and (\(2\)), we get:

\(\frac{AE}{EC} = \frac{AF}{FC}\)

Adding \(1\) to both sides, we get:

\(\frac{AE}{EC} + 1 = \frac{AF}{FC} + 1\)

\(\frac{AE + EC}{EC} = \frac{AF + FC}{FC}\)

\(\frac{AC}{EC} = \frac{AC}{FC}\)

\(EC = FC\) [Cancelling \(AC\)]

This is true only if \(F\) and \(E\) coincide.

Therefore, \(DE \parallel BC\)

Hence, we proved.