7. Quadrilateral $ABCD$ is inscribed in a circle with center $O$. Two circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii with centers $O_{1}$ and $O_{2}$ are inscribed in angles $BAD$ and $BCD$ respectively, with the first touching side $AD$ at point $K$, and the second touching side $BC$ at point $T$.

a) Find the radius of circle $\Omega_{1}$ if $AK=2$, $CT=8$.

b) Suppose it is additionally known that point $O_{2}$ is the center of the circle circumscribed around triangle $BOC$. Find the angle $BDC$.
Answer. a) $r=4$, b) $\angle B D C=\operatorname{arctg} \frac{\sqrt{5}-1}{2}$ or $\angle B D C=\pi-\operatorname{arctg} \frac{\sqrt{5}+1}{2}$.

Solution. a) Segments $A O_{1}$ and $C O_{2}$ are the bisectors of angles $B A D$ and $B C D$ (the center of the circle inscribed in an angle lies on the bisector of this angle). Since the quadrilateral $A B C D$ is inscribed in a circle, the sum of its opposite angles $B A D$ and $B C D$ is $180^{\circ}$, so the sum of their halves - angles $K A O_{1}$ and $T C O_{2}$ - is $90^{\circ}$. Let $\angle O_{1} A K=\alpha$. Then $\angle T O_{2} C=90^{\circ}-\angle T C O_{2}=\alpha$. Expressing $\operatorname{tg} \alpha$ in two ways, we get: $\operatorname{tg} \alpha=\frac{O_{1} K}{A K}=\frac{C T}{O_{2} T}, \frac{r}{2}=\frac{8}{r}, r=4$.

b) $O_{2} B=O_{2} C$ as radii of the circle circumscribed around triangle $B O C$, so the height of this triangle $O_{2} T$ is also its median. Points $O, O_{2}$, and $T$ lie on the same line (on the perpendicular bisector of segment $B C$). We then find: $O_{2} C=\sqrt{O_{2} T^{2}+C T^{2}}=4 \sqrt{5}, O_{2} O=O_{2} C=4 \sqrt{5}$ (as radii of the same circle).

There are two cases: points $O_{2}$ and $O$ can lie either on the same side of line $B C$ or on opposite sides of it.

In the first case, we get: $\angle B D C=\frac{1}{2} \angle B O C=\angle C O T=\operatorname{arctg} \frac{C T}{O T}=\operatorname{arctg} \frac{8}{4+4 \sqrt{5}}=\operatorname{arctg} \frac{\sqrt{5}-1}{2}$.

In the second case: $\angle B D C=\pi-\frac{1}{2} \angle B O C=\pi-\angle C O T=\pi-\operatorname{arctg} \frac{C T}{O T}=\pi-\operatorname{arctg} \frac{8}{4 \sqrt{5}-4}=\pi-\operatorname{arctg} \frac{\sqrt{5}+1}{2}$.