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:

Answer: Option A

Solution

Answer: Option A

Solution:

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