6. Examples

1. In a $\triangle ABC$, a straight line $DE$ intersects $AB$ at $D$ and $AC$ at $E$ and is parallel to $BC$, then prove that $\frac{AB}{AD} = \frac{AC}{AE}$

 

Solution:

 

 

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.

 

 

2. In a $\triangle ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively such that $\frac{AD}{DB} = \frac{AE}{EC}$ and $\angle ADE = \angle DEA$. Prove that $\triangle ABC$ is isosceles.

 

Solution:

 

 

Given that $\frac{AD}{DB} = \frac{AE}{EC}$, then by the converse of Thales theorem, we have:

 

$DE \parallel BC$

 

Therefore, $\angle ADE = \angle ABC$ [Corresponding angles] ---- ($1$)

 

and $\angle DEA = \angle BCA$ [Corresponding angles] ---- ($2$)

 

But, it is given that $\angle ADE = \angle DEA$ ---- ($3$)

 

Using ($1$) and ($2$) in equation ($3$), we get:

 

$\angle ABC = \angle BCA$

 

Therefore, $AC = AB$ [If opposite angles are equal, then opposite sides are equal.]

 

Thus, $\triangle ABC$ is isosceles.

 

Hence, we proved.