1. **Problem statement:** Given a circle with center $O$, diameter $AB$, and point $S$ on the circumference such that $\angle BOS = \theta$, with $OA = OB = r$ and $OD = x$. We need to apply the cosine rule to triangle $AOD$ to show that $$AD^2 = r^2 (1 + 3 \cos^2 \theta).$$ 2. **Recall the cosine rule:** For any triangle with sides $a$, $b$, and $c$, and angle $\gamma$ opposite side $c$, the cosine rule states: $$c^2 = a^2 + b^2 - 2ab \cos \gamma.$$ 3. **Identify sides and angle in $\triangle AOD$:** - $OA = r$ - $OD = x$ - $AD$ is the side opposite angle $\angle AOD = \theta$ 4. **Apply the cosine rule to $\triangle AOD$:** $$AD^2 = OA^2 + OD^2 - 2 \cdot OA \cdot OD \cos \theta = r^2 + x^2 - 2 r x \cos \theta.$$ 5. **Express $x$ in terms of $r$ and $\theta$:** Since $D$ lies on $OS$ and $BD$ is perpendicular to $OS$, using right triangle $BOD$ and the fact that $OB = r$, we find: $$BD \perp OS \implies BD \perp OD.$$ From the geometry, $BD = r \sin \theta$ and $OD = r \cos \theta$ (projection of $OB$ onto $OS$). Therefore, $$x = OD = r \cos \theta.$$ 6. **Substitute $x = r \cos \theta$ into the expression for $AD^2$:** $$AD^2 = r^2 + (r \cos \theta)^2 - 2 r (r \cos \theta) \cos \theta = r^2 + r^2 \cos^2 \theta - 2 r^2 \cos^2 \theta.$$ 7. **Simplify the expression:** $$AD^2 = r^2 + r^2 \cos^2 \theta - 2 r^2 \cos^2 \theta = r^2 - r^2 \cos^2 \theta = r^2 (1 - \cos^2 \theta).$$ 8. **Check the problem statement:** The problem states to show $$AD^2 = r^2 (1 + 3 \cos^2 \theta),$$ which differs from our current result. 9. **Re-examine the problem setup:** Since $AB$ is a diameter, $A$ and $B$ lie on the circle vertically, and $S$ lies on the circumference such that $\angle BOS = \theta$. Point $D$ is the foot of the perpendicular from $B$ to $OS$. 10. **Find $BD$ and $OD$ in terms of $r$ and $\theta$:** - $OB = r$ - $OS$ makes angle $\theta$ with $OB$ Since $BD$ is perpendicular to $OS$, in right triangle $BOD$: $$BD = r \sin \theta, \quad OD = r \cos \theta.$$ 11. **Find $AD$ using the Pythagorean theorem in triangle $ABD$:** Since $AB$ is diameter, $A$ is at $(0,r)$, $B$ at $(0,-r)$, and $O$ at origin. Coordinates: - $O = (0,0)$ - $B = (0,-r)$ - $S$ lies on circle, $OS$ makes angle $\theta$ with vertical axis. Vector $OS$ has direction $(\sin \theta, \cos \theta)$. Point $D$ lies on $OS$, so coordinates of $D$ are: $$D = t (\sin \theta, \cos \theta)$$ for some $t$. 12. **Find $t$ such that $BD \perp OS$:** Vector $BD = D - B = (t \sin \theta, t \cos \theta + r)$. Since $BD \perp OS$, their dot product is zero: $$(t \sin \theta, t \cos \theta + r) \cdot (\sin \theta, \cos \theta) = 0,$$ which gives $$t \sin^2 \theta + t \cos^2 \theta + r \cos \theta = 0,$$ $$t (\sin^2 \theta + \cos^2 \theta) + r \cos \theta = 0,$$ $$t + r \cos \theta = 0,$$ $$t = - r \cos \theta.$$ 13. **Coordinates of $D$:** $$D = (- r \cos \theta \sin \theta, - r \cos^2 \theta).$$ 14. **Coordinates of $A$:** $$A = (0, r).$$ 15. **Calculate $AD^2$:** $$AD^2 = (0 - (- r \cos \theta \sin \theta))^2 + (r - (- r \cos^2 \theta))^2 = (r \cos \theta \sin \theta)^2 + (r + r \cos^2 \theta)^2.$$ 16. **Simplify:** $$AD^2 = r^2 \cos^2 \theta \sin^2 \theta + r^2 (1 + \cos^2 \theta)^2 = r^2 \cos^2 \theta \sin^2 \theta + r^2 (1 + 2 \cos^2 \theta + \cos^4 \theta).$$ 17. **Use $\sin^2 \theta = 1 - \cos^2 \theta$:** $$AD^2 = r^2 \cos^2 \theta (1 - \cos^2 \theta) + r^2 (1 + 2 \cos^2 \theta + \cos^4 \theta) = r^2 (\cos^2 \theta - \cos^4 \theta + 1 + 2 \cos^2 \theta + \cos^4 \theta).$$ 18. **Combine like terms:** $$AD^2 = r^2 (1 + 3 \cos^2 \theta).$$ **Final answer:** $$\boxed{AD^2 = r^2 (1 + 3 \cos^2 \theta)}.$$