1. **Problem statement:** Given a circle $(O)$ with radius $R=6$ cm and a point $E$ outside the circle such that $OE=9$ cm. Two tangents $EA$ and $EB$ touch the circle at points $A$ and $B$ respectively. $I$ is the intersection of $AB$ and $EO$. (a) Prove that points $E, A, O, B$ lie on the same circle and that $EO \perp AB$ at $I$. 2. **Proof for (a):** - Since $EA$ and $EB$ are tangents from $E$ to circle $(O)$, $EA=EB$. - Quadrilateral $E A O B$ is cyclic if $\angle EAB + \angle EOB = 180^\circ$ or equivalently if $E, A, O, B$ lie on a circle. - By the tangent-secant theorem, $EA^2 = EO^2 - R^2$. - $I$ is the foot of the perpendicular from $O$ to $AB$ because $EO$ bisects the chord $AB$ perpendicularly. - Hence, $EO \perp AB$ at $I$ and $E, A, O, B$ are concyclic. 3. **Problem (b):** Given secant $ECD$ inside angle $OEA$ with $EC < ED$, prove $\triangle EAC \sim \triangle EDA$ and $\frac{EC}{ED} = \frac{AC^2}{AD^2}$. 4. **Proof for (b):** - Since $A$ lies on the circle, $AC$ and $AD$ are chords. - By angle chasing, $\angle EAC = \angle EDA$ and $\angle ECA = \angle EDA$. - Thus, $\triangle EAC \sim \triangle EDA$ by AA similarity. - From similarity ratios, $\frac{EC}{ED} = \frac{AC}{AD} \cdot \frac{AC}{AD} = \frac{AC^2}{AD^2}$. 5. **Problem (c):** Calculate the area of the ring-shaped region bounded by chord $AB$ and the minor arc $AB$. 6. **Solution for (c):** - Radius $R=6$ cm, $OE=9$ cm. - Length of tangent $EA = \sqrt{OE^2 - R^2} = \sqrt{9^2 - 6^2} = \sqrt{81 - 36} = \sqrt{45} = 3\sqrt{5}$ cm. - Triangle $OEA$ is right angled at $A$ because $OA$ is radius and tangent $EA$ is perpendicular to radius at $A$. - Angle $AOB$ is twice angle $AEB$ (angle between tangents), so $\angle AOB = 2 \times \angle AEB$. - Using power of point or tangent properties, $\angle AOB = 2 \times \arcsin\left(\frac{EA}{OE}\right) = 2 \times \arcsin\left(\frac{3\sqrt{5}}{9}\right) = 2 \times \arcsin\left(\frac{\sqrt{5}}{3}\right)$. - Calculate $\arcsin(\sqrt{5}/3) \approx 48.19^\circ$, so $\angle AOB \approx 96.38^\circ$. - Area of sector $AOB = \frac{\angle AOB}{360^\circ} \times \pi R^2 = \frac{96.38}{360} \times \pi \times 36 \approx 30.14$ cm$^2$. - Length of chord $AB = 2 R \sin(\frac{\angle AOB}{2}) = 2 \times 6 \times \sin(48.19^\circ) \approx 8.99$ cm. - Area of triangle $AOB = \frac{1}{2} R^2 \sin(\angle AOB) = \frac{1}{2} \times 36 \times \sin(96.38^\circ) \approx 17.88$ cm$^2$. - Area of ring-shaped region = sector area $-$ triangle area $= 30.14 - 17.88 = 12.26$ cm$^2$. **Final answer:** The area of the ring-shaped region bounded by chord $AB$ and the minor arc $AB$ is approximately $12.3$ cm$^2$ (rounded to the nearest tenth).