answer:**Step 1:** Connect points (B) and (D) to form (BD), and let (BD) intersect (AC) at point (O). **Step 2:** Observe the angles in the cyclic quadrilateral (ABCD): [ begin{aligned} angle ADB &= angle ACB, angle BAC &= angle BAO, angle ABD &= angle OBA. end{aligned} ] **Step 3:** This allows us to conclude that (triangle ABD) is similar to (triangle OBA) ((triangle ABD sim triangle OBA)) by AA similarity (Angle-Angle Similarity). Thus, we have: [ angle BOA = angle BAD = alpha. ] **Step 4:** Using the property of similar triangles, we know: [ AB^2 = BO cdot BD. ] **Step 5:** Apply the Ptolemy's theorem for cyclic quadrilateral (ABCD): [ AC cdot BD = AB cdot CD + BC cdot AD. ] Given (AB = BC = AD = CD), we can simplify the above expression as: [ AC cdot BD = AB(CD + AD) = AB^2 + AB^2 = 2AB^2. ] Since (AC = d) and (BD = AB cdot sqrt{2}), we have: [ d cdot AB cdot sqrt{2} = 2AB^2. ] Solving for (BO): [ BD = frac{d cdot sqrt{2}}{AB} implies BO = AC = d. ] **Step 6:** Now, calculate the area of (triangle ABC): [ S_{triangle ABC} = S_{triangle ABO} + S_{triangle OBC}. ] By the area formula for a triangle, we have: [ S_{triangle ABO} = frac{1}{2} AO cdot BO cdot sin(angle AOB), quad S_{triangle OBC} = frac{1}{2} OC cdot BO cdot sin(angle BOC). ] Since (angle AOB = angle BOC = alpha): [ S_{triangle ABC} = frac{1}{2} AC cdot BO cdot sin(alpha). ] Given that (AC = BO = d): [ S_{triangle ABC} = frac{1}{2} d^2 sin(alpha). ] **Conclusion:** The area of (triangle ABC) is: [ boxed{frac{1}{2} d^2 sin(alpha)} ]

answer:To prove that a polygon with an odd number of sides is regular if it is inscribed in a circle and all its sides are equal, we follow these steps: 1. **Properties of a Tangent Polygon:** Consider a polygon ( P ) with an odd number ( n ) of sides, all of which are equal, and inscribed in a circle ( Omega ). This means that each vertex of the polygon touches the circle. 2. **Tangents from Vertices:** Draw tangents from each vertex of the polygon to the circle ( Omega ). Let ( A_1, A_2, A_3, ldots, A_n ) be the vertices of the polygon. 3. **Equality of Tangents:** From each vertex ( A_i ), we draw tangents to the circle that touch the circle at points ( T_i ) and ( T_{i+1} ) respectively. Since the tangents from a given point to a circle to the point of tangency are equal, we have: [ A_iT_i = A_iT_{i+1} ] 4. **Midpoints of Sides:** Now consider any two consecutive vertices ( A_{i-1} ) and ( A_i ). The tangents ( A_{i-1}T_i ) and ( A_iT_i ) are equal in length. Because all the sides ( A_iA_{i+1} ) of the polygon are equal, each side of the polygon bisects the angle formed by the two tangents at each vertex. This implies that the points of tangency ( T_i ) are midpoints of the sides ( A_iA_{i+1} ). 5. **Symmetry and Regularity:** Since the polygon is symmetric about its center ( O ) and has an odd number of sides, the arguments about the tangents and midpoints can be extended to the entire polygon. This symmetry implies that each side is bisected by the radius drawn to the midpoint of every side. Consequently, all angles between consecutive vertices are equal, confirming that each internal angle is the same, adhering to the properties of a regular polygon. 6. **Conclusion:** Therefore, for a polygon inscribed in a circle with an odd number of equal sides, the points of tangency are the midpoints of the sides, and by the definition of a regular polygon, it is regular. Thus, we have shown that an inscribed polygon with all sides equal and an odd number of sides is indeed a regular polygon. (blacksquare)

answer:To find out how long Freddy takes to complete his journey, we first need to calculate the average speed of both Eddy and Freddy. Eddy's average speed can be calculated using the formula: Speed = Distance / Time For Eddy: Distance from city A to city B = 570 km Time taken by Eddy = 3 hours So, Eddy's average speed = 570 km / 3 hours = 190 km/h Now, we have the ratio of their average speeds: Eddy : Freddy = 2.533333333333333 Let's denote Freddy's average speed as F. Then we can write the ratio as: 190 km/h : F km/h = 2.533333333333333 To find F, we can set up the proportion: 190 / F = 2.533333333333333 Now, we solve for F: F = 190 / 2.533333333333333 F ≈ 75 km/h Now that we have Freddy's average speed, we can calculate the time he takes to travel from city A to city C using the formula: Time = Distance / Speed For Freddy: Distance from city A to city C = 300 km Freddy's average speed = 75 km/h So, the time taken by Freddy = 300 km / 75 km/h = 4 hours Therefore, Freddy takes boxed{4} hours to complete his journey from city A to city C.

answer:Let's assume the original length of the rectangle is ( L ) and the original width is ( W ). The original area of the rectangle, ( A_{original} ), is: [ A_{original} = L times W ] If each side is increased by 20%, the new length, ( L_{new} ), and the new width, ( W_{new} ), will be: [ L_{new} = L + 0.20L = 1.20L ] [ W_{new} = W + 0.20W = 1.20W ] The new area of the rectangle, ( A_{new} ), is: [ A_{new} = L_{new} times W_{new} ] [ A_{new} = (1.20L) times (1.20W) ] [ A_{new} = 1.20 times 1.20 times L times W ] [ A_{new} = 1.44 times L times W ] The percentage increase in the area is the change in area divided by the original area, multiplied by 100: [ text{Percentage Increase} = left( frac{A_{new} - A_{original}}{A_{original}} right) times 100 ] [ text{Percentage Increase} = left( frac{1.44LW - LW}{LW} right) times 100 ] [ text{Percentage Increase} = left( 1.44 - 1 right) times 100 ] [ text{Percentage Increase} = 0.44 times 100 ] [ text{Percentage Increase} = 44% ] So, the percentage increase in the area of the rectangle, if each of its sides is increased by 20%, is boxed{44%} .