ln a square ABCD, diagonals AC and BD interest at O. The angle bisector of ∠CAB meets BD and BC at F and G, respectively. OF : CG is equal to:
A. 1 : 2
B. 1 : 3
C. 1 : √3
D. 1 : √2
Answer: Option A
Solution(By Examveda Team)

$$\eqalign{ & ABCD{\text{ is a square}} \cr & AC = \sqrt 2 AB \cr & AO = OC = \frac{{AC}}{2} = \frac{{\sqrt 2 AB}}{2} = \frac{{AB}}{{\sqrt 2 }} \cr & \Delta AOF \sim \Delta ABG \cr & \left[ {{\text{By }}AA{\text{ property}}} \right] \cr & \frac{{AO}}{{AB}} = \frac{{OF}}{{BG}} \cr & \frac{{\frac{{AB}}{{\sqrt 2 }}}}{{AB}} = \frac{{OF}}{{BG}} \cr & \frac{1}{{\sqrt 2 }} = \frac{{OF}}{{BG}} \cr & BG = \sqrt 2 OF{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & AG{\text{ is angle bisector of }}\Delta ABC \cr & \frac{{AB}}{{AC}} = \frac{{BG}}{{GC}} = \frac{1}{{\sqrt 2 }} \cr & \left[ {{\text{angle bisector theorem}}} \right] \cr & BG = \frac{1}{{\sqrt 2 }}GC{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{Compare }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \sqrt 2 OF = \frac{1}{{\sqrt 2 }}GC \cr & OF:CG = 1:2 \cr} $$
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