In a triangle ABC,the mid-point of BC is D. I | vedclass

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(C) In $	riangle ACD$,since $AD \perp AC$,$\angle DAC = 90^\circ$. Thus,$\cos C = \frac{AC}{CD} = \frac{b}{a/2} = \frac{2b}{a}$.In $	riangle ABC$,by

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$\frac{2(c^2-a^2)}{3ac}$

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(C) In $	riangle ACD$,since $AD \perp AC$,$\angle DAC = 90^\circ$. Thus,$\cos C = \frac{AC}{CD} = \frac{b}{a/2} = \frac{2b}{a}$.In $	riangle ABC$,by the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Equating the two expressions for $\cos C$:$\frac{2b}{a} = \frac{a^2+b^2-c^2}{2ab}$$4b^2 = a^2+b^2-c^2$$3b^2 = a^2-c^2 \implies b^2 = \frac{a^2-c^2}{3}$.Now,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.$\cos A \cos C = \left(\frac{b^2+c^2-a^2}{2bc}ight) \left(\frac{2b}{a}ight) = \frac{b^2+c^2-a^2}{ac}$.Substituting $b^2 = \frac{a^2-c^2}{3}$:$\cos A \cos C = \frac{\frac{a^2-c^2}{3} + c^2 - a^2}{ac} = \frac{a^2-c^2+3c^2-3a^2}{3ac} = \frac{2c^2-2a^2}{3ac} = \frac{2(c^2-a^2)}{3ac}$.

In a $\triangle ABC$,the mid-point of $BC$ is $D$. If $AD$ is perpendicular to $AC$,then $\cos A \cos C=$

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