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(B) In $	riangle ABC$,$AD$ is the median to $BC$,so $BD = DC = a/2$. Given $AD \perp AC$,in right-angled $	riangle ADC$,by Pythagoras theorem,$AD^2

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$2c^2$

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(B) In $	riangle ABC$,$AD$ is the median to $BC$,so $BD = DC = a/2$. Given $AD \perp AC$,in right-angled $	riangle ADC$,by Pythagoras theorem,$AD^2 + b^2 = (a/2)^2$,so $AD^2 = a^2/4 - b^2$. By Apollonius theorem on $	riangle ABC$,$c^2 + b^2 = 2(AD^2 + (a/2)^2)$. Substituting $AD^2$,we get $c^2 + b^2 = 2(a^2/4 - b^2 + a^2/4) = a^2 - 2b^2$,which implies $a^2 = 3b^2 + c^2$ (Eq. $iii$). Now,consider the expression $3ca \cos A \cos C + 2a^2$. Using cosine rule,$\cos A = (b^2 + c^2 - a^2)/(2bc)$ and $\cos C = (a^2 + b^2 - c^2)/(2ab)$. Substituting these,the expression becomes $3ca \cdot [(b^2 + c^2 - a^2)/(2bc)] \cdot [(a^2 + b^2 - c^2)/(2ab)] + 2a^2$. $= (3/4b^2) \cdot (b^2 + c^2 - a^2)(a^2 + b^2 - c^2) + 2a^2$. Since $a^2 - c^2 = 3b^2$,we have $b^2 + c^2 - a^2 = -2b^2$ and $a^2 + b^2 - c^2 = 4b^2$. $= (3/4b^2) \cdot (-2b^2)(4b^2) + 2a^2 = -6b^2 + 2a^2 = 2(a^2 - 3b^2)$. From Eq. $iii$,$a^2 - 3b^2 = c^2$. Thus,the expression equals $2c^2$.

In $\triangle ABC$,if the median $AD$ drawn through $A$ is perpendicular to the side $AC$,then $3ca \cos A \cos C + 2a^2 =$

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