In the following \(▵ABC\), given that
\(DE ∥AB\).

We have to prove that \(\frac{DC}{AD}=\frac{EC}{BE} \)

Construction:
Let us join \(AE\) and \(BD\).

We also draw \(DP⊥EC\), and \(EQ⊥DC\).

In the above,

Area of \(△DEC = \frac12\times DC\times EQ\)

(\(DC\) as base, \(EQ\) as height)

Area of \(▵ADE = \frac12\times AD\times EQ\)

(\(AD\) as base, \(EQ\) as height)

\[\therefore \frac{\text{Area of }△DEC}{\text{Area of }▵ADE} = \frac{\frac12\times DC\times EQ}{\frac12\times AD\times EQ}\]
\[\therefore \frac{\text{Area of }△DEC}{\text{Area of }▵ADE} = \frac{DC}{AD}\quad\color{red} ...eqn. (i)\]

Again,

Area of \(△DEC = \frac12\times EC\times DP\)

(\(EC\) as base, \(DP\) as height)

Area of \(▵BDE = \frac12\times BE\times DP\)

(\(BE\) as base, \(DP\) as height)

\[\therefore \frac{\text{Area of }△DEC}{\text{Area of }▵BDE} = \frac{\frac12\times EC\times DP}{\frac12\times BE\times DP}\]
\[\therefore \frac{\text{Area of }△DEC}{\text{Area of }▵BDE} = \frac{EC}{BE}\quad\color{red} ...eqn. (ii)\]

Both \(△ADE\) and \(△BDE\) share the same base \(DE\), and they lie between the same parallel lines \(DE\) and \(AB\). So, \[\text{Area of }△ADE = \text{Area of }△BDE\] \[\therefore \frac{\text{Area of }△DEC}{\text{Area of }▵ADE} = \frac{\text{Area of }△DEC}{\text{Area of }▵BDE}\quad\color{red} ...eqn. (iii)\] From \(\color{red} (i) \) \(\color{red} (ii) \) and \(\color{red} (iii) \), \[\frac{DC}{AD} = \frac{EC}{BE} \quad\color{green}\text{Proved}\]

The above theorem can also be proved by using the conception of similarity of triangles.

Since \(DE \) and \(AB\) are parallel lines,

In \(△ABC\) and \(△DEC\),

\(∠CAB = ∠CDE\quad\) [corresponding angles],

\(∠CBA = ∠CED\quad\) [corresponding angles],

\(∠ACB = ∠DCE\quad\) [common angles].

So, by \(AA\) rule, \(△ABC ∼ △DEC\).

Because corresponding sides of similar tringles are proportional,

\begin{align}
&\frac{AC}{DC}=\frac{BC}{EC} \\\\
&\Rightarrow \frac{DC+AD}{DC}=\frac{EC+BE}{EC} \\\\
&\Rightarrow \frac{DC}{DC} + \frac{AD}{DC}=\frac{EC}{EC} + \frac{BE}{EC} \\\\
&\Rightarrow 1 + \frac{AD}{DC}=1 + \frac{BE}{EC} \\\\
&\Rightarrow \frac{AD}{DC}=\frac{BE}{EC} \quad\color{gray}\text{[ cancelling 1 from both sides]} \\\\
&\therefore \frac{DC}{AD}=\frac{EC}{BE} \quad\color{gray}\text{[ taking reciprocals]} \\\\
\end{align}

The Converse of Basic proportionality theorem states that
If a line divides any two sides of a triangle in the same ratio,
then the line is parallel to the third side.

In the following \(▵ABC\), the line \(DE\) intersects \(CA\) at point \(D\),
and \(CB\) at point \(E\) in such a way, that \(\frac{DC}{AD} = \frac{EC}{BE}\).

We have to prove that \(DE ∥ AB\).

Construction: We draw a line \(DF\) parallel to \(AB\).

According to Basic proportionality theorem,

\begin{align}
&\frac{DC}{AD} = \frac{FC}{BF} \\\\
&\Rightarrow\frac{EC}{BE} = \frac{FC}{BF} \quad\color{gray}\left[\because \;\small\frac{DC}{AD} = \frac{EC}{BE}\;\textsf{given}\right]\\\\
&\Rightarrow 1 + \frac{EC}{BE} = 1 + \frac{FC}{BF} \quad\color{gray}\textsf{[adding 1 to both sides]}\\\\
&\Rightarrow\frac{BE+EC}{BE} = \frac{BF+FC}{BF} \\\\
&\Rightarrow\frac{BC}{BE} = \frac{BC}{BF} \quad\color{gray}[\small\because BE+CE=BC,\; BF+FC=BC]\\\\
&\therefore BE = BF \quad\color{gray}\textsf{[BC cancelled from both sides]}
\end{align}

∴ It is clear that
points \(E\) and \(F\) are coincident.

∴ \(DE ∥ AB\) ∵ \(DF ∥ AB\).

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