Let ABC be a triangle such that AB = BC. Let | vedclass

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(C) Let $BC = 4x$. Since $BX = 3XC$ and $BC = BX + XC$,we have $BX = 3x$ and $XC = x$.Given $AB = BC$,so $AB = 4x$.Since $F$ is the mid-point of $AB$,

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$\sqrt{\frac{3}{2}}$

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(C) Let $BC = 4x$. Since $BX = 3XC$ and $BC = BX + XC$,we have $BX = 3x$ and $XC = x$.Given $AB = BC$,so $AB = 4x$.Since $F$ is the mid-point of $AB$,$BF = AF = 2x$.In $	riangle BFX$,$\angle BFX = 90^\circ$. Thus,$\cos B = \frac{BF}{BX} = \frac{2x}{3x} = \frac{2}{3}$.In $	riangle ABC$,by the Law of Cosines:$\cos B = \frac{AB^2 + BC^2 - AC^2}{2(AB)(BC)}$$\frac{2}{3} = \frac{(4x)^2 + (4x)^2 - AC^2}{2(4x)(4x)}$$\frac{2}{3} = \frac{32x^2 - AC^2}{32x^2}$$64x^2 = 3(32x^2 - AC^2)$$64x^2 = 96x^2 - 3AC^2$$3AC^2 = 32x^2$$AC^2 = \frac{32x^2}{3}$$AC = \sqrt{\frac{32}{3}}x = 4x \sqrt{\frac{2}{3}}$Therefore,$\frac{BC}{AC} = \frac{4x}{4x \sqrt{\frac{2}{3}}} = \sqrt{\frac{3}{2}}$.

Let $ABC$ be a triangle such that $AB = BC$. Let $F$ be the mid-point of $AB$ and $X$ be a point on $BC$ such that $FX$ is perpendicular to $AB$. If $BX = 3XC$,then the ratio $BC / AC$ equals

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