answer:Since arc CD is 45^circ, angle CAD = frac{45^circ}{2} = 22.5^circ. As angle AOD = 180^circ - angle CAD = 180^circ - 22.5^circ = 157.5^circ. Triangle AOC is isosceles with AO = CO, so angle OAC = angle OCA = 22.5^circ. Since angle ABO = 45^circ, this is an exterior angle to triangle BCO, thus angle BOC = angle ABO - angle BCO = 45^circ - 22.5^circ = 22.5^circ. As a result, triangle BOC is isosceles with BO = OC. Using the Law of Sines: [ frac{BC}{sin(angle BOC)} = frac{BO}{sin(angle BCO)} ] [ frac{BC}{sin(22.5^circ)} = frac{8}{sin(22.5^circ)} ] [ BC = 8 ] Thus, AC = BC = boxed{8}.

answer:1. **Understanding the Given Problem** We need to prove that there is a unique perpendicular segment whose endpoints lie on two skew lines. Two lines are said to be skew if they do not intersect and are not parallel. 2. **Key Concept and Approach** We will leverage the property that through a line and a point not on the line, there exists a unique perpendicular line. Here we consider lines ( l_1 ) and ( l_2 ), which are given as skew lines. Our goal is to show that there is a unique segment perpendicular to both lines with endpoints on these lines. 3. **Step-by-Step Proof** - **Step 3.1: Consideration of Line ( l )** Consider a line ( l ) that is perpendicular to both ( l_1 ) and ( l_2 ). We need to construct such a line and then prove its uniqueness. This can be done by creating planes through each line that are parallel to ( l ). - **Step 3.2: Constructing Parallel Plane** Through ( l_1 ), draw a plane ( P_1 ) which is parallel to the line ( l ). Since ( l ) is perpendicular to ( l_1 ), the plane constructed will intersect ( l_2 ) at some point. - **Step 3.3: Intersection Point** Let the intersection point of the plane ( P_1 ) and line ( l_2 ) be ( A ). This ensures the segment ( overline{AB} ) is the one we are looking for, where ( A in l_1 ) and ( B in l_2 ). - **Step 3.4: Projection Analysis** To confirm these points form the ends of the unique perpendicular, consider projecting ( l_1 ) and ( l_2 ) onto a plane parallel to ( l ). The intersection of these projections will give us the exact points for endpoints of the unique perpendicular segment. 4. **Conclusion: Ensuring Uniqueness** Given that the construction of the plane and the projection leads to distinct and unique points for the segments ends, we conclude that such a perpendicular segment is both existent and unique. Hence, we have proven the required statement. [ boxed{} ]

answer:1. Since (x, y, z) form a harmonic progression, we know that (frac{1}{x}, frac{1}{y}, frac{1}{z}) form an arithmetic progression. This implies that: [ frac{2}{y} = frac{1}{x} + frac{1}{z} ] 2. Rearranging and simplifying the equation: [ frac{2}{y} = frac{x+z}{xz} implies 2xz = y(x+z) ] 3. Let us consider the expression given in the problem: [ lg (x+z) + lg (x - 2y + z) ] 4. Using the logarithm property ( lg a + lg b = lg(ab) ): [ lg (x+z) + lg (x - 2y + z) = lg left[(x+z)(x - 2y + z)right] ] 5. Expanding the product inside the logarithm: [ (x+z)(x - 2y + z) = x(x - 2y + z) + z(x - 2y + z) ] [ = x^2 - 2xy + xz + zx - 2yz + z^2 ] [ = x^2 - 2xy + xz + xz - 2yz + z^2 ] [ = x^2 - 2xy + 2xz - 2yz + z^2 ] 6. Simplifying further using the relation (2xz = y(x+z)): [ x^2 - 2xy + 2xz - 2yz + z^2 = (x^2 - 2xz + z^2) + 2xz - 2xy - 2yz ] 7. Factorizing the terms, we notice: [ x^2 - 2xz + z^2 = (x - z)^2 ] [ 2yz = 2xz~text{(because}~2xz = y(x+z) implies 2yz=2xztext{)} ] 8. Hence: [ (x+z)(x - 2y + z) = (x - z)^2 ] 9. Therefore: [ lg left[(x+z)(x - 2y + z)right] = lg left[(x - z)^2right] ] 10. Using the logarithm property (lg (a^2) = 2 lg a): [ lg left[(x - z)^2right] = 2 lg (x - z) ] 11. Thus, we have proven that: [ lg (x+z) + lg (x-2 y+z) = 2 lg (x-z) ] Conclusion: [ boxed{} ]

answer:-1 **Key Point:** Properties of odd and even functions. **Topic:** Properties and applications of functions. **Analysis:** First, based on f(x+2)=-f(x), we get f(x)=-f(x-2). Therefore, f(7) can be transformed into -f(1)=-1. From f(x+2)=-f(x), we get: f(x)=-f(x-2); Therefore, f(7)=-f(5)=f(3)=-f(1)=-[2(1)-1]=boxed{-1}. **Review:** This problem examines the ability to deduce f(x)=-f(x-2) from f(x+2)=-f(x), and to know that to find f(7), we need to change the value of the variable 7 to fall within the interval [0,1].