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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.$\frac{\overline{AC }+ \overline{CB}}{\overline{AB}} = \frac{\sqrt{2}r +\sqrt{2}r }{2r} = \frac{2\sqrt{2}r }{2r} = \frac{\sqrt{2} }{1}=\sqrt{2}$

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