In a square ABCD, diagonals AC and BD intersect at O. The angle bisector of ∠CAB meets BD and BC at F and G, respectively. OF ∶ CG is equal to:

Given:

ABCD is a square

Diagonal intersects at O

The angle bisector of ∠CAB meets BD and BC at F and G

Calculation:

Let the side AB be x 

So, Diagonal AC = √2.x

⇒ OA = OB = OC = OD = AC/2 = x/√2    ----(i)

Now, In ΔAOB and from angle bisector theorem 

AB : AO = BF : OF

⇒ x : (x/√2)  = BF/OF

⇒ √2 : 1 = BF/OF

Let BF = √2y, OF = y      ----(ii)

Since, BF + OF = OB = x/√2     [From (i)]

⇒ √2y + y = x/√2

⇒ \(y = \frac{x}{\sqrt2 . (\sqrt2 + 1)}\)

⇒ \(OF = \frac{x}{\sqrt2 . (\sqrt2 + 1)}\)    ----(iii)        [from (ii)]

Now, In ΔABC and from angle bisector theorem 

AB : AC = BG : GC

⇒ x : √2x  = BG : GC

⇒ 1 : √2 = BG : GC

Let BG = z, GC = √2z      ----(iv)

Since, BG + GC = BC = x   

⇒ z + √2z = x 

⇒ \(z = \frac{x}{(\sqrt2 + 1)}\)

⇒ \(\sqrt2.z = \frac{\sqrt2x}{(\sqrt2 + 1)}\)

⇒ \(CG = \frac{\sqrt2x}{(\sqrt2 + 1)}\)      ----(v)

Now from (iii) and (v), we get 

OF : CG = \( \frac{x}{\sqrt2 . (\sqrt2 + 1)} : \frac{\sqrt2x}{(\sqrt2 + 1)}\)

⇒ OF : CG = \( \frac{1}{\sqrt2} : \frac{\sqrt2}{1}\)

⇒ OF : CG = 1 : 2

∴ The required ratio is 1 : 2.