5. Theorems

 

Given: In a triangle $ABC$, a straight line $l$ parallel to $BC$ intersects $AB$ at $D$ and $AC$ at $E$.

 

To prove: $\frac{AD}{DB} = \frac{AE}{EC}$

 

Construction: Join $BE$ and $CD$. Draw $EF \perp AB$ and $DG \perp CA$.

 

Proof: Since $EF \perp AB$, $EF$ is the height of $\triangle ADE$ and $\triangle DBE$.

 

$ar(\triangle ADE) = \frac{1}{2} \times AD \times EF$ and

 

$ar(\triangle DBE) = \frac{1}{2} \times DB \times EF$

 

Therefore, $\frac{ar(\triangle ADE)}{ar(\triangle DBE)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} = \frac{AD}{DB}$ ---- ($1$)

 

Similarly, $ar(\triangle ADE) = \frac{1}{2} \times AE \times DG$ and

 

$ar(\triangle DCE) = \frac{1}{2} \times EC \times DG$

 

Thus, $\frac{ar(\triangle ADE)}{ar(\triangle DCE)} = \frac{\frac{1}{2} \times AE \times DG}{\frac{1}{2} \times EC \times DG} = \frac{AE}{EC}$ ---- ($2$)

 

But, $\triangle DBE$ and $\triangle DCE$ are on the same base and between the same parallels $BC$ and $DE$.

 

Therefore, $ar(\triangle DBE) = ar(\triangle DCE)$ ---- ($3$)

 

From ($1$), ($2$) and ($3$), we have:

 

$\frac{AD}{DB} = \frac{AE}{EC}$

 

Hence, we proved.

 

Now, we shall learn the Converse of Basic proportionality theorem or Converse of Thales theorem.

 

Given: A line $l$ intersects the sides $AB$ and $AC$ of $\triangle ABC$ at $D$ and $E$, respectively, such that $\frac{AD}{DB} = \frac{AE}{EC}$ ---- ($1$)

 

To prove: $DE \parallel BC$

 

Construction: If $DE$ is not parallel to $BC$, then draw a line $DF \parallel BC$.

 

Proof: Since $DF \parallel BC$, by Thales theorem, we get:

 

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

 

From ($1$) and ($2$), we have:

 

$\frac{AF}{FC} = \frac{AE}{EC}$

 

Add $1$ to both sides of the equation, we have:

 

$\frac{AF}{FC} + 1 = \frac{AE}{EC} + 1$

 

$\frac{AF + FC}{FC} = \frac{AE + EC}{EC}$

 

$\frac{AC}{FC} = \frac{AC}{EC}$

 

$FC = EC$

 

This is true only if $F$ and $E$ coincide.

 

Therefore, $DE \parallel BC$

 

Hence, we proved.