question_answer

BE, CF are the two medians of \[\Delta ABC\] and G is their point of intersection. EF cuts AG at O. The ratio of AO and OG is equal to                                                                                              [SSC (10+2) 2015]

A) 1 : 3                             

B) 2 : 3

C) 3 : 1     

D) 1 : 2

Correct Answer: C

Solution :

F and E are mid-points of AB and AC, respectively

So,       \[EF\parallel BC\]

Also in \[\Delta ADB,\]     \[FO\parallel BD\]

\[\Rightarrow \]   \[\frac{AF}{FB}=\frac{AO}{OD}\]

\[\Rightarrow \]   \[\frac{AF}{FB}+1=\frac{AO}{OD}+1\]

\[\Rightarrow \]   \[\frac{AF+FB}{FB}=\frac{OA+OD}{OD}\]

\[\Rightarrow \]   \[\frac{AB}{FB}=\frac{AD}{OD}\]\[\Rightarrow \]\[\frac{2FB}{FB}=\frac{AD}{OD}\]

So, O is mid-point of AD.

\[\therefore \]      \[OA=OD\]

\[\Rightarrow \]   \[OA=OG+DG\]

\[\Rightarrow \]   \[OA=OG+\frac{AG}{2}\]                     \[\left( \because \,\,\frac{AG}{GD}=\frac{2}{1} \right)\]

\[\Rightarrow \]   \[OA=OG+\frac{AO+OG}{2}\]

\[\Rightarrow \]   \[2OA=3\,\,OG+AO\]\[\Rightarrow \]\[\frac{AO}{AG}=\frac{3}{1}=3:1\]