In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC.

Solution:

(i) It is given that $\triangle A B C$ and $D E \| B C$

We have to find the $A C$

Since

$A D=6 \mathrm{~cm}$

$D B=9 \mathrm{~cm}$

$A E=8 \mathrm{~cm}$

$\Rightarrow A B=15$

So $\frac{A D}{B D}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{6}{9}=\frac{8}{x}$

$6 x=72 \mathrm{~cm}$

$x=\frac{72}{6} \mathrm{~cm}$

$=12 \mathrm{~cm}$

Hence

$\begin{aligned} A C &=12+8 \\ &=20 \end{aligned}$

(ii)lt is given that $\frac{A D}{B D}=\frac{3}{4}$ and $A C=15 \mathrm{~cm}$

We have to find $A E$

Let $A E=x$

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{3}{4}=\frac{x}{15-x}$

$45-3 x=4 x$

$-3 x-4 x=-45$

$7 x=45$

$x=\frac{45}{7}$

Hence

$x=6.43 \mathrm{~cm}$

(iii)lt is given that $\frac{A D}{B D}=\frac{2}{3}$ and $A C=18 \mathrm{~cm}$

We have to find $A E$

Let $A E=x$ and $C E=18-x$

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{2}{3}=\frac{x}{18-x}$

$3 x=36-2 x$

 

$5 x=36 \mathrm{~cm}$

$x=\frac{36}{5} \mathrm{~cm}$

$x=7.2 \mathrm{~cm}$

Hence

$A E=7.2 \mathrm{~cm}$

(iv)lt is given that $A D=4 \mathrm{~cm}, A E=8 \mathrm{~cm}, D B=x-4$ and $E C=3 x-19$.

We have to find $x$

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{4}{x-4}=\frac{8}{3 x-19}$

$4(3 x-19)=8(x-4)$

$12 x-76=8(x-4)$

$12 x-8 x=-32+76$

$4 x=44 \mathrm{~cm}$

Hence

$x=11 \mathrm{~cm}$

(v) It is given that $A D=8 \mathrm{~cm}, A B=12 \mathrm{~cm}$ and $A E=12 \mathrm{~cm}$.

We have to find CE.

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{8}{4}=\frac{12}{C E}$

$8 C E=4 \times 12 \mathrm{~cm}$

$C E=\frac{4 \times 12}{8} \mathrm{~cm}$

$=\frac{48}{8} \mathrm{~cm}$

$=6 \mathrm{~cm}$

Hence

$C E=6 \mathrm{~cm}$

(vi) It is given that $A D=4 \mathrm{~cm}, D B=4.5 \mathrm{~cm}$ and $A E=8 \mathrm{~cm}$.

We have to find $A C$.

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{4}{4.5}=\frac{8}{A C}$

$A C=\frac{4.5 \times 8}{4} \mathrm{~cm}$

$=9 \mathrm{~cm}$

Hence

$A C=9 \mathrm{~cm}$

(vii) It is given that $A D=2 \mathrm{~cm}, A B=6 \mathrm{~cm}$ and $A C=9 \mathrm{~cm}$.

We have to find $A E$.

Now

DB = 6 − 2 = 4 cm

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{2}{4}=\frac{x}{9-x} \quad($ Let $x=A E)$

$4 x=18-2 x$

 

$6 x=18 \mathrm{~cm}$

$x=\frac{18}{6} \mathrm{~cm}$

$x=3 \mathrm{~cm}$

Hence

$x=3 \mathrm{~cm}$

(viii) It is given that $\frac{A D}{B D}=\frac{4}{5}$ and $E C=2.5 \mathrm{~cm}$

We have to find $A E$.

So $\frac{A D}{D B}=\frac{A E}{C E}$  (by Thales theorem)

Then $\frac{4}{5}=\frac{A E}{2.5}$

$\mathrm{AE}=4 \times 2.55=2 \mathrm{~cm}$

Hence

$A E=2 \mathrm{~cm}$

(ix) It is given that $A D=x, D B=x-2, A E=x+2$ and $E C=x-1$.

We have to find the value of $x$.

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{x}{x-2}=\frac{x+2}{x-1}$

$x(x-1)=(x-2)(x+2)$

$x^{2}-x-x^{2}+4=0$

$x=4$

Hence

$x=4 \mathrm{~cm}$

$(\mathrm{x})$ It is given that $A D=8 x-7, D B=5 x-3, A E=4 x-3$ and $E C=3 x-1$

We have to find the value of $x$.

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

Then,

$8 x-75 x-3=4 x-33 x-1 \Rightarrow 8 x-73 x-1=5 x-34 x-3 \Rightarrow 24 x 2-29 x+7=20 \times 2-27 x+9 \Rightarrow 4 \times 2-2 x-2=0 \Rightarrow 22 \times 2-x-1=0 \Rightarrow 2 \times 2-x-1=0 \Rightarrow 2 x 2-2 x+x-1=0 \Rightarrow 2 x x-1+1 x-1=0 \Rightarrow x-12 x+1=0 \Rightarrow x-$

 

$1=0$ or $2 x+1=0 \Rightarrow x=1$ or $x=-12$ rejected

Hence,

$x=1 \mathrm{~cm}$

(xi) It is given that $A D=4 x-3, B D=3 x-1, A E=8 x-7$ and $E C=5 x-3$.

We have to find the value of $x$.

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{4 x-3}{3 x-1}=\frac{8 x-7}{5 x-3}$

4x-35x-3=3x-18x-7

$4 x(5 x-3)-3(5 x-3)=3 x(8 x-7)-1(8 x-7)$

$20 x^{2}-12 x-15 x+9=24 x^{2}-21 x-8 x+7$

$20 x^{2}-27 x+9=24 x^{2}-29 x+7$

Then

$-4 x^{2}+2 x+2=0$

 

$4 x^{2}-2 x-2=0$

$4 x^{2}-4 x+2 x-2=0$

 

$4 x(x-1)+2(x-1)=0$

$(4 x+2)(x-1)=0$

 

$x=1$

Hence

$x=1 \mathrm{~cm}$

(xii) It is given that $A D=2.5 \mathrm{~cm}, A E=3.75 \mathrm{~cm}$ and $B D=3 \mathrm{~cm}$.

So $\frac{A D}{D B}=\frac{A E}{C E}$ (by Thales theorem)

Then $\frac{2.5}{3}=\frac{3.75}{C E}$

$2.5 C E=3.75 \times 3$

$C E=\frac{3.75 \times 3}{2.5}$

$=\frac{11.25}{2.5}$

$=4.50$

Now

$A C=3.75 \mathrm{~cm}+4.50 \mathrm{~cm}$

$=8.25 \mathrm{~cm}$