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In \[\Delta ABC,\] \[DE\parallel BC\]where D is a point on AB and E is a point on AC. DE divides the area of \[\Delta \,ABC\]into two equal parts. Then, DB: AB is equal to [SSC CGL Tier II, 2015]

A) \[\sqrt{2}:(\sqrt{2}-1)\]

B) \[(\sqrt{2}+1):\sqrt{2}\]

C) \[(\sqrt{2}-1):\sqrt{2}\]

D) \[\sqrt{2}:(\sqrt{2}+1)\]

Correct Answer: C

Solution :

[c]                                     \[\Delta ADE\] is Similar to\[\Delta \,ABC\] and \[\frac{\text{Area}\,\,\text{o}f\,\,\Delta ADE}{\text{Area}\,\,\text{of}\,\,\Delta ABC}\]                         \[=\frac{1}{2}\]\[\Rightarrow \]\[\frac{AD}{AB}=\frac{1}{\sqrt{2}}\]             \[\therefore \]      \[\frac{DB}{AB}=1-\frac{DB}{AB}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}\]