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. |
\[\therefore \] \[OA=OD\] |
\[\Rightarrow \] \[OA=OG+DG\] |
\[\Rightarrow \] \[OA=OG+\frac{AG}{2}\] \[\left( \because \,\,\frac{AG}{GD}=\frac{2}{1} \right)\] |
\[\Rightarrow \] \[OA=OG+\frac{AO+OG}{2}\] |
\[\Rightarrow \] \[2OA=3\,\,OG+AO\]\[\Rightarrow \]\[\frac{AO}{AG}=\frac{3}{1}=3:1\] |
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