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
Question:

In a $\triangle A B C, D$ and $E$ are points on the sides $A B$ and $A C$ respectively. For each of the following cases show that $D E \| B C$ :

(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2  and AE = 1.8 cm.
(iii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.
(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm

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

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.