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
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
(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.