Basic Proportionality Theorem of Triangles

If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the other two sides are divided in the same ratio.

Basic proportionality theorem: proof

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 \(\color{red} (iii) \), \[\frac{DC}{AD} = \frac{EC}{BE} \quad\color{green}\text{Proved}\]

Basic proportionality theorem by similarity of triangles

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}

Converse of Basic proportionality theorem

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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