Correct Answer: $1: \sqrt{2}$

Solution :
Given: ABCD is a square. Diagonal intersects at O. The angle bisector of $\angle$CAB meets BD and BC at F and G.
Let's denote the side AB as \(x\). So, the diagonal AC = \(\sqrt{2}x\).
$⇒OA = OB = OC = OD = \frac{AC}{2} = \frac{x}{\sqrt{2}}$ ___(i)
Now, in $\triangle$AOB and from the angle bisector theorem,
We have \(\frac{AB}{AO} = \frac{BF}{OF}\)
\(⇒\frac{x}{\frac{x}{\sqrt{2}}} = \frac{BF}{OF}\)
Let's denote BF as \(\sqrt{2}y\) and OF as \(y\). From equation (i),
We know that \(BF + OF = OB = \frac{x}{\sqrt{2}}\)
\(⇒\sqrt{2}y + y = \frac{x}{\sqrt{2}}\)
\(⇒y = \frac{x}{\sqrt{2} + 2}\)
\(⇒OF = \frac{x}{\sqrt{2} + 2}\) ___ (iii)
In $\triangle$ABC and from the angle bisector theorem,
We have \(\frac{AB}{AC} = \frac{BG}{GC}\)
\(⇒\frac{x}{\sqrt{2}x} = \frac{BG}{GC}\)
Let's denote BG as \(z\) and GC as \(\sqrt{2}z\).___ (iv)
Since \(BG + GC = BC = x\)
We have \(z + \sqrt{2}z = x\)
\(⇒z = \frac{x}{\sqrt{2} + 1}\)
\(⇒CG = \frac{2x}{\sqrt{2} + 1}\)___ (v)
Now from equations (iii) and (v),
We get \(OF : CG = \frac{x}{\sqrt{2} + 2} : \frac{2x}{\sqrt{2} + 1}\)
\(OF : CG = 1 : 2\).
Hence, the correct answer is $1:2$.