Question : In $△$ ABC, D and E are points on AB and AC, respectively, such that DE || BC and DE divide the $△$ ABC into two parts of equal areas. The ratio of AD and BD is:
Option 1: $1 : 1$
Option 2: $1 : (\sqrt{2} -$ 1)
Option 3: $1 : \sqrt{2}$
Option 4: $1 : (\sqrt{2} +$ 1)

Question

Question : $D$ and $E$ are points on the sides $A B$ and $A C$ respectively of $△ A B C$ such that $D E$ is parallel to $B C$ and $A D : D B = 4 : 5$ , $C D$ and $B E$ intersect each other at $F$ . Find the ratio of the areas of $△ D E F$ and $△ C B F$ .
Option 1: $16 : 25$
Option 2: $16 : 81$
Option 3: $81 : 16$
Option 4: $4 : 9$

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Answer 1

Correct Answer: $1 : (\sqrt{2} -$ 1)

Solution :
Area( $△$ ADE) = Area(trapezium BCED)
⇒ Area( $△$ ADE) + Area( $△$ ADE) = Area(trapezium BCED) + Area( $△$ ADE)
⇒ 2 × Area( $△$ ADE) = Area( $△$ ABC)
In $△$ ADE and $△$ ABC,
$∠$ ADE = $∠$ ABC [i.e., corresponding angles in DE || BC]
$∠$ A = $∠$ A [common angle]
⇒ $△$ ADE ~ $△$ ABC
⇒ $\frac{Area(△ ADE)}{Area(△ ABC)}$ = $\frac{A D^{2}}{A B^{2}}$
⇒ $\frac{Area(△ ADE)}{2×Area(△ ADE)}$ = $\frac{A D^{2}}{A B^{2}}$
⇒ $\frac{1}{2}$ = $\frac{A D^{2}}{A B^{2}}$
⇒ $\frac{A D}{A B}$ = $\frac{1}{\sqrt{2}}$ ------------------(i)
⇒ AB = $\sqrt{2}$ AD
⇒ AD + BD = $\sqrt{2}$ AD
⇒ BD = $(\sqrt{2} - 1)$ AD
⇒ AD : BD = $1 : (\sqrt{2} - 1)$
Hence, the correct answer is $1 : (\sqrt{2} - 1)$ .

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Question : In △ABC, D and E are points on AB and AC, respectively, such that DE || BC and DE divide the △ABC into two parts of equal areas. The ratio of AD and BD is:Option 1: 1:1Option 2: 1:(2−1)Option 3: 1:2Option 4: 1:(2+1)

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Question : In $△$ ABC, D and E are points on AB and AC, respectively, such that DE || BC and DE divide the $△$ ABC into two parts of equal areas. The ratio of AD and BD is:
Option 1: $1 : 1$
Option 2: $1 : (\sqrt{2} -$ 1)
Option 3: $1 : \sqrt{2}$
Option 4: $1 : (\sqrt{2} +$ 1)