Ans A
Given O as the center. AB is the diameter. Let Diameter = $2r$
Then $ AO = OB = r $, $ CO = r $ Semicircle.
$\angle COB = \angle COA= 90^{\circ}$, Applying Pythagoras Theorem in $\triangle COB, \triangle COA$
$CB = CA = \sqrt{2}r$
Putting in given equation.$\frac{\overline{AC }+ \overline{CB}}{\overline{AB}} = \frac{\sqrt{2}r +\sqrt{2}r }{2r} = \frac{2\sqrt{2}r }{2r} = \frac{\sqrt{2} }{1}=\sqrt{2}$