In a triangle ABC,the angle bisector BD of an | vedclass

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(B) Let $AB = c$ and $AD = x$. By the Angle Bisector Theorem,$\frac{AB}{BC} = \frac{AD}{CD}$,so $\frac{c}{2} = \frac{x}{1}$,which implies $x = \frac{c

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$\frac{15}{2}$

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(B) Let $AB = c$ and $AD = x$. By the Angle Bisector Theorem,$\frac{AB}{BC} = \frac{AD}{CD}$,so $\frac{c}{2} = \frac{x}{1}$,which implies $x = \frac{c}{2}$.In $	riangle BCD$,by the Law of Cosines,$BD^2 = BC^2 + CD^2 - 2(BC)(CD) \cos C$.$\left(\frac{3}{\sqrt{2}}ight)^2 = 2^2 + 1^2 - 2(2)(1) \cos C \implies \frac{9}{2} = 5 - 4 \cos C \implies 4 \cos C = \frac{1}{2} \implies \cos C = \frac{1}{8}$.In $	riangle ABC$,by the Law of Cosines,$c^2 = AB^2 = BC^2 + AC^2 - 2(BC)(AC) \cos C$.$c^2 = 2^2 + (x+1)^2 - 2(2)(x+1) \left(\frac{1}{8}ight)$.Substituting $x = \frac{c}{2}$,$c^2 = 4 + (\frac{c}{2}+1)^2 - (\frac{c}{2}+1) = 4 + \frac{c^2}{4} + c + 1 - \frac{c}{2} - 1 = \frac{c^2}{4} + \frac{c}{2} + 4$.$\frac{3c^2}{4} - \frac{c}{2} - 4 = 0 \implies 3c^2 - 2c - 16 = 0$.$(3c - 8)(c + 2) = 0$. Since $c > 0$,$c = \frac{8}{3}$.Then $AD = x = \frac{c}{2} = \frac{4}{3}$.Perimeter $= AB + BC + AC = c + 2 + (x+1) = \frac{8}{3} + 2 + \frac{4}{3} + 1 = 4 + 3 = 7$. Wait,re-evaluating the provided solution logic: The formula $BD = \frac{2ac}{a+c} \cos \frac{B}{2}$ is standard. Using $a=2, CD=1, AD=x, c=AB$,we have $x/1 = c/2 \implies x = c/2$. $AC = c/2 + 1$. Using Stewart's Theorem on $	riangle ABC$ with cevian $BD$: $c^2(1) + 2^2(c/2) = (c/2+1)(BD^2 + (c/2)(1))$.$c^2 + 2c = (c/2+1)(9/2 + c/2) = (c/2+1)(c+9)/2 = (c^2 + 9c + 2c + 18)/4$.$4c^2 + 8c = c^2 + 11c + 18 \implies 3c^2 - 3c - 18 = 0 \implies c^2 - c - 6 = 0 \implies (c-3)(c+2) = 0$. So $c=3$.Then $AD = c/2 = 3/2$. Perimeter $= 3 + 2 + (3/2 + 1) = 5 + 2.5 = 7.5 = \frac{15}{2}$.

In a $\triangle ABC$,the angle bisector $BD$ of $\angle B$ intersects $AC$ in $D$. Suppose $BC=2$,$CD=1$ and $BD=\frac{3}{\sqrt{2}}$. The perimeter of the $\triangle ABC$ is

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