In a triangle ABC,it is known that AB=AC. Sup | vedclass

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(C) Let $AD = CD = x$. Then $AC = 2x$ and $AB = 2x$ (since $AB = AC$).In $	riangle BCD$,by the Law of Cosines:$BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos C$$

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$\sqrt{7}$

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(C) Let $AD = CD = x$. Then $AC = 2x$ and $AB = 2x$ (since $AB = AC$).In $	riangle BCD$,by the Law of Cosines:$BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos C$$2^2 = 2^2 + x^2 - 2(2)(x)\cos C$$4 = 4 + x^2 - 4x\cos C \Rightarrow \cos C = \frac{x}{4}$.Since $AB = AC$,$\angle B = \angle C$,so $\cos B = \cos C = \frac{x}{4}$.In $	riangle ABC$,$\angle A = 180^\circ - 2C$. Thus,$\cos A = \cos(180^\circ - 2C) = -\cos(2C) = -(2\cos^2 C - 1) = 1 - 2(\frac{x^2}{16}) = 1 - \frac{x^2}{8}$.Using the Law of Cosines in $	riangle ABC$:$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos A$$2^2 = (2x)^2 + (2x)^2 - 2(2x)(2x)(1 - \frac{x^2}{8})$$4 = 4x^2 + 4x^2 - 8x^2(1 - \frac{x^2}{8})$$4 = 8x^2 - 8x^2 + x^4$$x^4 = 4$ $\Rightarrow x^2 = 2$ $\Rightarrow x = \sqrt{2}$.Thus,$AC = 2\sqrt{2}$ and $BC = 2$.$\cos C = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}$.$\sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \frac{1}{8}} = \sqrt{\frac{7}{8}} = \frac{\sqrt{7}}{2\sqrt{2}}$.Area of $	riangle ABC = \frac{1}{2} 	imes AC 	imes BC 	imes \sin C = \frac{1}{2} 	imes 2\sqrt{2} 	imes 2 	imes \frac{\sqrt{7}}{2\sqrt{2}} = \sqrt{7}$.

In a $\triangle ABC$,it is known that $AB=AC$. Suppose $D$ is the mid-point of $AC$ and $BD=BC=2$. Then,the area of the $\triangle ABC$ is

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