6. A triangle $\triangle A B C$ has its vertices lying on a circle $\mathrm{C}$ of radius 1 , with $\angle B A C=60^{\circ}$. A circle with center $I$ is inscribed in $\triangle A B C$. The line $A I$ meets circle $\mathrm{C}$ again at $D$. Find the length of the segment $I D$.
6. Ans: 1
$A D$ bisects the angle $\angle A$ and $I C$ bisects the angle $\angle C$. Now $\angle B C D=\angle B A D=\angle A / 2$. $\angle A D C=\angle B$. Hence $\angle I C D=(\angle A+\angle C) / 2$ and $\angle D I C=180^{\circ}-\angle B-(\angle A+\angle C) / 2=$ $(\angle A+\angle C) / 2$. Thus $I D=C D$. The chord $C D$ subtends an angle of $30^{\circ}$ at point $A$ of circle $\mathbb{C}$. Hence it subtends an angle of $60^{\circ}$ at the center of circle $\mathbb{C}$. Thus $I D=C D=2 \sin \left(60^{\circ} / 2\right)=1$.