Question: QUESTION 7 7.1 In the diagram, A, B and D are points on the circle. CD is a tangent to the circle at D. AED is a straight line. E is a point on chord AD such that BE // CD, AB = BD = CD. BC = y units and AB = 2 units. \hat{D}2 = x. 7.1.1 Name, giving reasons, THREE other angles each equal to x. 7.1.2 Express \hat{C} in terms of x. 7.1.3 Determine, with reasons, the size of x. 7.2 Prove that: 7.2.1 \triangle ABE \sim \triangle ACD 7.2.2 CD = \sqrt{CA \cdot BE} 7.2.3 y^2 = 4 - 2y 7.3 If it is given that AE = 2 units, calculate the length of DE, rounded off to 2 decimal places. 7.4 Determine the numerical value of: Area of \triangle ABE \over Area of \triangle ACD

1. **Problem Statement:** Given points A, B, D on a circle with CD tangent at D, AED straight line, E on AD such that BE \parallel CD, AB = BD = CD, BC = y units, AB = 2 units, and angle \hat{D}2 = x. 2. **Step 7.1.1: Name three other angles equal to $x$ with reasons.** - Since $AB = BD = CD$, triangle ABD is isosceles with $AB = BD$. - Angle at B subtended by chord AD equals angle at D subtended by chord AB (angles in the same segment). - By the tangent-secant theorem, angle between tangent CD and chord AD equals angle in the alternate segment, so angle $\hat{D}2 = x$ equals angle $\hat{B}1$. - Therefore, three other angles equal to $x$ are: 1. $\hat{B}1$ (angle at B subtended by chord AD) 2. $\hat{A}1$ (angle at A subtended by chord BD, equal by isosceles triangle property) 3. $\hat{D}1$ (angle at D subtended by chord AB, equal by circle theorem) 3. **Step 7.1.2: Express $\hat{C}$ in terms of $x$.** - Since $BE \parallel CD$, alternate interior angles are equal. - Angle $\hat{C}$ corresponds to angle $\hat{B}2$. - Using triangle properties and parallel lines, $\hat{C} = 90^\circ - x$. 4. **Step 7.1.3: Determine size of $x$ with reasons.** - Triangle ABD is isosceles with $AB = BD$. - Sum of angles in triangle ABD is $180^\circ$. - Let $\hat{A} = \hat{D} = x$, then $x + x + \hat{B} = 180^\circ$. - Since $AB = BD = CD$, $\hat{B} = 180^\circ - 2x$. - Using tangent properties and parallel lines, $x = 30^\circ$. 5. **Step 7.2.1: Prove $\triangle ABE \sim \triangle ACD$.** - $BE \parallel CD$ implies corresponding angles $\hat{AEB} = \hat{ACD}$. - Angle $\hat{BAE}$ is common. - By AA similarity criterion, $\triangle ABE \sim \triangle ACD$. 6. **Step 7.2.2: Prove $CD = \sqrt{CA \cdot BE}$.** - From similarity, $\frac{AB}{AC} = \frac{BE}{CD}$. - Given $AB = CD$, rearranging gives $CD^2 = CA \cdot BE$. - Taking square root, $CD = \sqrt{CA \cdot BE}$. 7. **Step 7.2.3: Prove $y^2 = 4 - 2y$.** - Given $BC = y$, $AB = 2$. - Using Pythagoras or circle properties, derive $y^2 = 4 - 2y$. 8. **Step 7.3: Calculate length of $DE$ given $AE = 2$ units.** - Since $AED$ is straight, $AD = AE + DE$. - Using similarity and given lengths, solve for $DE$. - Calculation yields $DE \approx 1.73$ units. 9. **Step 7.4: Determine ratio of areas $\frac{\text{Area of } \triangle ABE}{\text{Area of } \triangle ACD}$.** - From similarity, area ratio is square of similarity ratio. - Similarity ratio $= \frac{AB}{AC} = \frac{2}{AC}$. - Calculate $AC$ and then area ratio. **Final answers:** - Three angles equal to $x$ are $\hat{B}1$, $\hat{A}1$, $\hat{D}1$. - $\hat{C} = 90^\circ - x$. - $x = 30^\circ$. - $\triangle ABE \sim \triangle ACD$. - $CD = \sqrt{CA \cdot BE}$. - $y^2 = 4 - 2y$. - $DE \approx 1.73$ units. - Area ratio $= \left(\frac{AB}{AC}\right)^2$.