In a ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC

Solution:

(i) It is given that $D$ and $E$ are point on sides $A B$ and $A C$.

We have to prove that DE || BC.

According to Thales theorem we have

$\frac{A D}{D B}=\frac{A E}{C E}$

$\Rightarrow \frac{8}{4}=\frac{12}{6}$

$\Rightarrow 2=2 \quad$ (Proportional)

Hence, DE || BC.

(ii) It is given that $D$ and $E$ are point on sides $A B$ and $A C$.

We have to prove that DE || BC.

According to Thales theorem we have

$\frac{A D}{D B}=\frac{A E}{C E}$

$\Rightarrow \frac{1.4}{4.2}=\frac{1.8}{5.4}$

$\Rightarrow \frac{1}{3}=\frac{1}{3} \quad$ (Proportional)

Hence, DE || BC.

(iii) It is given that $D$ and $E$ are point on sides $A B$ and $A C$.

We have to prove that DE || BC.

According to Thales theorem we have

$\frac{A D}{D B}=\frac{A E}{C E}$

So 

AD=AB-DB=10.8-4.5=6.3

And 

EC=AC-AE=4.8-2.8=2

Now

$\frac{6.3}{4.5}=\frac{2.8}{2.0}$

Hence, DE || BC.

(iv) It is given that $D$ and $E$ are point on sides $A B$ and $A C$.

We have to prove that DE || BC.

According to Thales theorem we have

$\frac{A D}{D B}=\frac{A E}{C E}$

$\Rightarrow \frac{5.7}{9.5}=\frac{3.3}{5.5}$

$\Rightarrow \frac{3}{5}=\frac{3}{5} \quad$ (Proportional)

Hence, DE || BC.