$D$ and $E$ are points on the sides $A B$ and $A C$ respectively of a $\triangle A B C$ such that $D E \| B C$. Find the value of $x$, when
(i) $A D=x \mathrm{~cm}, D B=(x-2) \mathrm{cm}$,
$A E=(x+2) \mathrm{cm}$ and $E C=(x-1) \mathrm{cm}$.
(ii) $A D=4 \mathrm{~cm}, D B=(x-4) \mathrm{cm}, A E=8 \mathrm{~cm}$ and $E C=(3 x-19) \mathrm{cm}$.
(iii) $A D=(7 x-4) \mathrm{cm}, A E=(5 x-2) \mathrm{cm}$, $D B=(3 x+4) \mathrm{cm}$ and $E C=3 x \mathrm{~cm}$.
(i)
In $\triangle A B C$, it is given that $D E \| B C$.
Applying Thales' theorem, we have :
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
$\Rightarrow \frac{x}{x-2}=\frac{x+2}{x-1}$
$\Rightarrow x(x-1)=(x-2)(x+2)$
$\Rightarrow x^{2}-x=x^{2}-4$
$\Rightarrow x=4 \mathrm{~cm}$
(ii)
In $\triangle \mathrm{ABC}$, it is given that $\mathrm{DE} \| \mathrm{BC}$.
Applying Thales' theorem, we have :
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
$\Rightarrow \frac{4}{\mathrm{x}-4}=\frac{8}{3 \mathrm{x}-19}$
$\Rightarrow 4(3 \mathrm{x}-19)=8(\mathrm{x}-4)$
$\Rightarrow 12 \mathrm{x}-76=8 \mathrm{x}-32$
$\Rightarrow 4 \mathrm{x}=44$
$\Rightarrow \mathrm{x}=11 \mathrm{~cm}$
(iii)
In $\triangle \mathrm{ABC}$, it is given that $\mathrm{DE} \| \mathrm{BC}$.
Applying Thales' theorem, we have:
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
$\Rightarrow \frac{7 \mathrm{x}-4}{3 \mathrm{x}+4}=\frac{5 \mathrm{x}-2}{3 \mathrm{x}}$
$\Rightarrow 3 \mathrm{x}(7 \mathrm{x}-4)=(5 \mathrm{x}-2)(3 \mathrm{x}+4)$
$\Rightarrow 21 \mathrm{x}^{2}-12 \mathrm{x}=15 \mathrm{x}^{2}+14 \mathrm{x}-8$
$\Rightarrow 6 \mathrm{x}^{2}-26 \mathrm{x}+8=0$
$\Rightarrow(\mathrm{x}-4)(6 \mathrm{x}-2)=0$
$\Rightarrow \mathrm{x}=4, \frac{1}{3}$
$\because \mathrm{x} \neq \frac{1}{3} \quad$ (as if $x=\frac{1}{3}$ then $A E$ will become negative)
$\therefore \mathrm{x}=4 \mathrm{~cm}$