In an isosceles triangle ABC,angle C = angle | vedclass

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(C) Let $I$ be the incenter of $	riangle ABC$ and $BD$ be the median to side $AC$. Since $AB = BC$,the median $BD$ is also the angle bisector of $\an

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$3$

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(C) Let $I$ be the incenter of $	riangle ABC$ and $BD$ be the median to side $AC$. Since $AB = BC$,the median $BD$ is also the angle bisector of $\angle B$ and the altitude to $AC$.In $	riangle ABC$,$I$ lies on $BD$. The distance $ID = r$,where $r$ is the inradius.In $	riangle ABD$,$\angle BDA = 90^{\circ}$,so $BD = AD \cot \frac{B}{2}$.Also,$ID = r = (s-b) 	an \frac{B}{2}$.Given that $I$ divides the median $BD$ in the ratio $3:1$ from $B$ to $D$,we have $\frac{BI}{ID} = \frac{3}{1}$.Since $BD = BI + ID$,we have $BI = 3ID = 3r$.In $	riangle ABD$,$BD = BI + ID = 3r + r = 4r$.Using the property $r = (s-b) 	an \frac{B}{2}$ and $BD = \frac{2 \Delta}{b} = \frac{2 \sqrt{s(s-a)(s-b)(s-c)}}{b}$,for an isosceles triangle with $a=c$,$BD = \sqrt{a^2 - (b/2)^2}$.Alternatively,using $BI = r \csc \frac{B}{2}$,we have $\frac{BI}{ID} = \frac{r \csc \frac{B}{2}}{r} = \csc \frac{B}{2}$.Given $\frac{BI}{ID} = 3$,therefore $\csc \frac{B}{2} = 3$.

In an isosceles triangle $ABC$,$\angle C = \angle A$. If the point of intersection of the bisectors of internal angles $\angle A$ and $\angle C$ divides the median of side $AC$ in the ratio $3 : 1$ (from vertex $B$ to side $AC$),then the value of $\csc \frac{B}{2}$ is equal to

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