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In the given ...

Question

In the given figure, ABC is a triangle, DE is parallel to BC and $\frac{A D}{D B} = \frac{3}{2}$
(i) Determine the ratios $\frac{A D}{A B} a n d \frac{D E}{B C} .$
(ii) Prove that $Δ D E F$ is similar to $Δ C B F$ Hence, find $\frac{E F}{F B}$ .
(iii) What is the ratio of the areas of $Δ D F E a n d Δ B F C$ ? [4 MARKS]

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Solution

Application of theorems: 2 Marks
Calculation: 2 Marks

Proof: In $Δ A D E$ and $Δ A B C$ ,
$\begin{matrix} ∠ A = ∠ A & [c o m m o n] ∠ 1 = ∠ 2 & [C o r r e s p o n d i n g a n g l e s] S o, & Δ A D E \sim Δ A B C & [A A s i m i l a r i t y] H e n c e & \frac{A D}{A B} = \frac{D E}{B C} = \frac{A E}{A C} (i) G i v e n & \frac{A D}{D B} = \frac{3}{2} L e t & A D = 3 x . D B = 2 x S o, & \frac{A D}{A B} = \frac{A D}{A D + D B} = \frac{3 x}{3 x + 2 x} = \frac{3 x}{5 x} = \frac{3}{5} N o w, & \frac{A D}{A B} = \frac{D E}{B C} S o, & \frac{D E}{B C} = \frac{3}{5} \end{matrix}$

(ii) In $Δ D E F$ and $Δ C B F$ ,
$\begin{matrix} ∠ 3 = ∠ 4 & [V e r t i c a l l y o p p o s i t e a n g l e s] ∠ 5 = ∠ 6 & [A l t e r n a t e i n t e r i o r a n g l e s] S o, & Δ D E F \sim Δ C B F & [A A s i m i l a r i t y] H e n c e & \frac{D E}{B C} = \frac{E F}{B F} = \frac{D F}{C F} N o w, & \frac{D E}{B C} = \frac{E F}{F B} S o, & \frac{E F}{F B} = \frac{3}{5} & [∵ \frac{D E}{B C} = \frac{3}{5}] (i i i) & \frac{a r (Δ D E F)}{a r (Δ B C F)} = \frac{D E^{2}}{B C^{2}} = {(\frac{3}{5})}^{2} = \frac{9}{25} \end{matrix}$

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