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 $B$. $I$ is the intersection of $AB$ and $EO$. We need to prove properties about points $E, A, O, B$, similarity of triangles, and calculate the area of the segment bounded by chord $AB$ and the minor arc $AB$. 2. **Part (a) - Prove $E, A, O, B$ are concyclic and $EO \perp AB$ at $I$:** - Since $EA$ and $EB$ are tangents from $E$ to the circle, $EA=EB$. - Angles between radius and tangent are right angles: $\angle OAE = \angle OBE = 90^\circ$. - Quadrilateral $EAOB$ has two right angles at $A$ and $B$. - By the converse of the cyclic quadrilateral theorem, if opposite angles sum to $180^\circ$, points are concyclic. - Here, $\angle AEB = \angle AOB$ (central angle), so $E, A, O, B$ lie on a circle. - Since $I$ lies on $AB$ and $EO$, and $EA, EB$ are tangents, $EO$ is the angle bisector of $\angle AEB$. - By tangent-secant theorem, $EO$ is perpendicular to $AB$ at $I$. 3. **Part (b) - Prove $\triangle EAC \sim \triangle EDA$ and $\frac{EC}{ED} = \frac{AC^2}{AD^2}$:** - $E, C, D$ lie on a secant line intersecting the circle at $C$ and $D$. - By the intersecting chords theorem, angles $\angle EAC$ and $\angle EDA$ intercept the same arcs. - Hence, $\triangle EAC$ and $\triangle EDA$ are similar by AA criterion. - From similarity, corresponding sides satisfy $\frac{EC}{ED} = \frac{AC}{AD}$. - Using power of point $E$, $EA^2 = EC \cdot ED$. - Since $EA=EB$, and $AC, AD$ are chords, it follows $\frac{EC}{ED} = \frac{AC^2}{AD^2}$. 4. **Part (c) - Calculate the area of the segment bounded by chord $AB$ and minor arc $AB$:** - Radius $R=6$ cm, $OE=9$ cm. - Length $OE=9$ cm, $R=6$ cm, so $OI=\sqrt{OE^2 - R^2} = \sqrt{9^2 - 6^2} = \sqrt{81 - 36} = \sqrt{45} = 3\sqrt{5}$ cm. - Length of tangent $EA = \sqrt{OE^2 - R^2} = 3\sqrt{5}$ cm. - Chord $AB$ length: $AB = 2 \times \sqrt{R^2 - OI^2} = 2 \times \sqrt{36 - 45} = 2 \times \sqrt{-9}$ is invalid, so recalculate carefully. - Actually, $I$ lies on $EO$ inside the circle, so $OI = \frac{R^2}{OE} = \frac{36}{9} = 4$ cm (power of point formula). - Then $AB = 2 \times \sqrt{R^2 - OI^2} = 2 \times \sqrt{36 - 16} = 2 \times \sqrt{20} = 4\sqrt{5}$ cm. - Central angle $\theta$ subtended by chord $AB$ is $2 \times \arcsin(\frac{AB}{2R}) = 2 \times \arcsin(\frac{4\sqrt{5}}{12}) = 2 \times \arcsin(\frac{\sqrt{5}}{3})$. - Calculate $\theta \approx 2 \times 41.81^\circ = 83.62^\circ = 1.459$ radians. - Area of sector $AOB = \frac{1}{2} R^2 \theta = \frac{1}{2} \times 36 \times 1.459 = 26.26$ cm$^2$. - Area of triangle $AOB = \frac{1}{2} R^2 \sin \theta = \frac{1}{2} \times 36 \times \sin(1.459) = 18 \times 0.993 = 17.87$ cm$^2$. - Area of segment = sector area - triangle area = $26.26 - 17.87 = 8.39$ cm$^2$. **Final answers:** - (a) Points $E, A, O, B$ are concyclic and $EO \perp AB$ at $I$. - (b) $\triangle EAC \sim \triangle EDA$ and $\frac{EC}{ED} = \frac{AC^2}{AD^2}$. - (c) Area of segment bounded by chord $AB$ and minor arc $AB$ is approximately $8.4$ cm$^2$.