# D and E are points on the sides AB and AC, respectively, of a

Question:

$D$ and $E$ are points on the sides $A B$ and $A C$, respectively, of a $\triangle A B C$, such that $D E \| B C$.

(i) If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.
(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.

(iii) If $\frac{A D}{D B}=\frac{4}{7}$ and $A C=6.6 \mathrm{~cm}$, find $A E$.

(iv) If $\frac{A D}{A B}=\frac{8}{15}$ and $E C=3.5 \mathrm{~cm}$, find $A E$.

Solution:

(i)

In $\triangle A B C$, it is given that $D E \| B C$.

Applying Thales' theorem, we get:

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

$\because \mathrm{AD}=3.6 \mathrm{~cm}, \mathrm{AB}=10 \mathrm{~cm}, \mathrm{AE}=4.5 \mathrm{~cm}$

$\therefore \mathrm{DB}=10-3.6=6.4 \mathrm{~cm}$

or, $\frac{3.6}{6.4}=\frac{4.5}{E C}$

or, $E C=\frac{6.4 \times 4.5}{3.6}$

$o r, E C=8 \mathrm{~cm}$

Thus, $A C=A E+E C$

$=4.5+8=12.5 \mathrm{~cm}$

(ii)

In $\triangle \mathrm{ABC}$, it is given that $\mathrm{DE} \| \mathrm{BC}$.

Applying Thales' Theorem, we get:

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

Adding 1 to both sides, we get:

$\frac{\mathrm{AD}}{\mathrm{DB}}+1=\frac{\mathrm{AE}}{\mathrm{EC}}+1$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{DB}}=\frac{\mathrm{AC}}{\mathrm{EC}}$

$\Rightarrow \frac{13.3}{\mathrm{DB}}=\frac{11.9}{5.1}$

$\Rightarrow \mathrm{DB}=\frac{13.3 \times 5.1}{11.9}=5.7 \mathrm{~cm}$

Therefore, $A D=A B-D B=13.5-5.7=7.6 \mathrm{~cm}$

(iii)

In $\triangle A B C$, it is given that $D E \| B C$.

Applying Thales' theorem, we get:

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

$\Rightarrow \frac{4}{7}=\frac{\mathrm{AE}}{\mathrm{EC}}$

Adding 1 to both the sides, we get:

$\frac{11}{7}=\frac{\mathrm{AC}}{\mathrm{EC}}$

$\Rightarrow \mathrm{EC}=\frac{6.6 \times 7}{11}=4.2 \mathrm{~cm}$

Therefore,

$\mathrm{AE}=\mathrm{AC}-\mathrm{EC}=6.6-4.2=2.4 \mathrm{~cm}$

(iv)

In $\triangle \mathrm{ABC}$, it is given that $\mathrm{DE} \| \mathrm{BC}$.

Applying Thales' theorem, we get:

$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$

$\Rightarrow \frac{8}{15}=\frac{\mathrm{AE}}{\mathrm{AE}+\mathrm{EC}}$

$\Rightarrow \frac{8}{15}=\frac{\mathrm{AE}}{\mathrm{AE}+3.5}$

$\Rightarrow 8 \mathrm{AE}+28=15 \mathrm{AE}$

$\Rightarrow 7 \mathrm{AE}=28$

$\Rightarrow \mathrm{AE}=4 \mathrm{~cm}$