Question : $D$ and $E$ are points of the sides $AB$ and $AC$, respectively of $\triangle ABC$ such that $DE$ is parallel to $BC$ and $AD:DB=7: 9$. If $CD$ and $BE$ intersect each other at $F$, then find the ratio of areas of $\triangle DEF$ and $\triangle CBF$.
Option 1: 49 : 144
Option 2: 49 : 81
Option 3: 49 : 256
Option 4: 256 : 49
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Correct Answer: 49 : 256
Solution : Since $\angle A$ is common and $DE||BC$, $\triangle ADE\sim\triangle ABC$ And given: $AD:DB=7:9$ ⇒ $AD:(AD+DB) = 7:(7+9)$ $\therefore AD:(AB)= 7:16$ Since $\triangle ADE\sim\triangle ABC$, $AD:AB = DE:BC = AE:AC = 7:16$ Since $DE||BC$, $\angle FDE= \angle FCB$ (alternate interior angles) $\angle FED = \angle FBC$ (alternate interior angles) And $\angle DFE = \angle CFB$ (vertically opposite angles) So, $\triangle DEF\sim\triangle CBF$ by AAA criteria Now, the area of similar triangles is proportional to squares of sides of triangles. So, Area ($\triangle DEF$) : Area ($\triangle CBF$) = $DE^2:BC^2$ ⇒ Area($\triangle DEF$) : Area ($\triangle CBF$) = $7^2 : 16^2$ $\therefore$ Area ($\triangle DEF$) : Area($\triangle CBF$) = 49 : 256 Hence, the correct answer is 49 : 256.
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