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(C) In $	riangle OAC$,$OA = 1$ and $AC:CO = 2:1$,so $AC = \frac{2}{3}$ and $OC = \frac{1}{3}$.In $	riangle OCD$,$OD = 1$ (radius) and $OC = \frac{1}

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$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(C) In $	riangle OAC$,$OA = 1$ and $AC:CO = 2:1$,so $AC = \frac{2}{3}$ and $OC = \frac{1}{3}$.In $	riangle OCD$,$OD = 1$ (radius) and $OC = \frac{1}{3}$. By Pythagoras theorem,$CD = \sqrt{OD^2 - OC^2} = \sqrt{1^2 - (\frac{1}{3})^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.In $	riangle ACD$,$AD = \sqrt{AC^2 + CD^2} = \sqrt{(\frac{2}{3})^2 + (\frac{2\sqrt{2}}{3})^2} = \sqrt{\frac{4}{9} + \frac{8}{9}} = \sqrt{\frac{12}{9}} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.Since $OE \perp AD$,in $	riangle OAC$ and $	riangle OEH$,$\angle OAC = \angle OEH = 90^\circ$ is not correct,rather $	riangle OEH \sim 	riangle ACD$ is not directly applicable. Instead,use coordinates: $O(0,0)$,$A(-1,0)$,$C(-\frac{1}{3}, 0)$,$D(-\frac{1}{3}, \frac{2\sqrt{2}}{3})$.The line $AD$ passes through $(-1,0)$ and $(-\frac{1}{3}, \frac{2\sqrt{2}}{3})$. Slope $m = \frac{\frac{2\sqrt{2}}{3} - 0}{-\frac{1}{3} - (-1)} = \frac{2\sqrt{2}/3}{2/3} = \sqrt{2}$.Equation of $AD$: $y - 0 = \sqrt{2}(x + 1) \Rightarrow y = \sqrt{2}x + \sqrt{2}$.Line $OE$ is perpendicular to $AD$ and passes through $(0,0)$,so its slope is $-\frac{1}{\sqrt{2}}$.Equation of $OE$: $y = -\frac{1}{\sqrt{2}}x$.Intersection $H$ of $CD$ $(x = -\frac{1}{3})$ and $OE$ $(y = -\frac{1}{\sqrt{2}}x)$: $y_H = -\frac{1}{\sqrt{2}}(-\frac{1}{3}) = \frac{1}{3\sqrt{2}}$.$D$ has $y_D = \frac{2\sqrt{2}}{3} = \frac{4}{3\sqrt{2}}$.$DH = y_D - y_H = \frac{4}{3\sqrt{2}} - \frac{1}{3\sqrt{2}} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$.

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