answer:Let's begin by interpreting the problem and using the given information to solve it step by step. Step 1: Understanding the Geometric Setup - We are given a triangle ( triangle ABC ). - A line ( DE ) is parallel to ( BC ) and intersects the line segments ( AB ) and ( AC ) at points ( D ) and ( E ), respectively. - Let ( P ) be an arbitrary point on ( DE ). - The lines ( BP ) and ( CP ) intersect ( AC ) at ( N ) and ( AB ) at ( M ) respectively. - We are given ( frac{AD}{DB} = lambda ), and need to show that ( lambda = lambda_1 + lambda_2 ), where ( lambda_1 = frac{AM}{MB} ) and ( lambda_2 = frac{AN}{NC} ). # Step 2: Using the Area Ratios Since ( DE parallel BC ), triangles ( triangle ADE ) and ( triangle ABC ) are similar by AA similarity. The ratio of their corresponding sides will hence be the same. The line ( DE ) divides the triangle ( triangle ABC ) into two similar triangles: ( triangle ADE ) and ( triangle ABC ). Hence, their areas are related by the square of the ratio of their sides because the triangles are similar: [ frac{[ADE]}{[ABC]} = left( frac{AD}{AB} right)^2 = left( frac{lambda}{lambda + 1} right)^2. ] # Step 3: Expressing Areas in Terms of Similarity Let’s compute the necessary area ratios: [ frac{[AMB]}{[ABC]} + frac{[ANC]}{[ABC]}. ] Given: [ frac{[AMB]}{[ABC]} = frac{AM}{AB} cdot frac{MB}{AB} quad text{and} quad frac{[ANC]}{[ABC]} = frac{AN}{AC} cdot frac{NC}{AC}. ] # Step 4: Using Area Properties Since ( DE parallel BC ), ( D ) and ( E ) divide ( AB ) and ( AC ) respectively in the ratio ( lambda : 1 ). Using properties of similar triangles: [ frac{AD}{DB} = lambda = frac{AE}{EC}. ] # Step 5: Applying Mass Points Using mass points, if ( D ) divides ( AB ) in the ratio ( lambda: 1 ), we can assign mass ( 1 cdot lambda ) to point ( A ) and mass ( 1 ) to point ( B ). Therefore, the total mass at point ( D ) will be ( lambda + 1 ). Using similar reasoning, the mass at ( E ) will be ( 1 cdot lambda ) plus mass at ( C ), which is ( 1 ). # Step 6: Connecting via Barycentric Coordinates Thus, in ( triangle ABP ): [ frac{AM}{MB} = lambda_1. ] In ( triangle ACP, ) comparing the ratios: [ frac{AN}{NC} = lambda_2. ] Since the line ( DE ) (parallel to ( BC )) divides the triangle area exactly: [ boxed{ lambda = lambda_1 + lambda_2 }.

answer:1. Given that ABCD is a convex quadrilateral with m(widehat{ADB}) = 15^circ and m(widehat{BCD}) = 90^circ, and the diagonals intersect perpendicularly at E. 2. We are given the lengths |EC| = 4, |EA| = 8, and |EP| = 2. 3. We need to find m(widehat{PBD}). First, let's use the given information about the lengths and the angles to find the necessary relationships. 4. Since m(widehat{BCD}) = 90^circ, triangle BCD is a right triangle with E as the foot of the altitude from B to CD. 5. The diagonals intersect perpendicularly at E, so BE cdot DE = CE^2 (this is a property of the right triangle where the altitude to the hypotenuse creates two similar triangles). 6. Given |EC| = 4, we have: [ BE cdot DE = 4^2 = 16 ] 7. Since |EA| = 8 and |EP| = 2, we can use the Power of a Point theorem at point P: [ PE cdot EA = 2 cdot 8 = 16 ] 8. This implies that triangle BEP sim triangle AED by the AA (Angle-Angle) similarity criterion because they share the angle at E and the right angle at E. 9. From the similarity, we have: [ angle PBE = angle DAE ] 10. Given m(widehat{ADB}) = 15^circ, we know that: [ angle DAE = 90^circ - 15^circ = 75^circ ] 11. Therefore, angle PBE = 75^circ. 12. Finally, since angle PBD = angle PBE - angle EBD and angle EBD = 0^circ (as E lies on BD), we have: [ angle PBD = 75^circ ] Conclusion: [ boxed{75^circ} ]

answer:First, let's find out how many turns each wheel makes in one minute. Wheel A makes 6 turns every 30 seconds, so in one minute (which is 60 seconds), it will make: 6 turns / 30 seconds = x turns / 60 seconds x = (6 turns * 60 seconds) / 30 seconds x = 360 turns / 30 x = 12 turns per minute Wheel B makes 10 turns every 45 seconds, so in one minute, it will make: 10 turns / 45 seconds = y turns / 60 seconds y = (10 turns * 60 seconds) / 45 seconds y = 600 turns / 45 y = 13.333... turns per minute (we'll keep the repeating decimal for now and round at the end if necessary) Now, let's find out how many turns each wheel makes in two hours. There are 60 minutes in an hour, so two hours have 120 minutes. For Wheel A: 12 turns/minute * 120 minutes = 1440 turns in two hours For Wheel B: 13.333... turns/minute * 120 minutes = 1600 turns in two hours (rounded to the nearest whole number) Now, let's add the total turns made by both wheels in two hours: Wheel A: 1440 turns Wheel B: 1600 turns Total turns = 1440 + 1600 = 3040 turns Both wheels make a total of boxed{3040} turns in two hours.

answer:To solve this, we note that translating the graph of the function y=sin 2x to the right by dfrac{pi}{12} units changes the function to y=sin left[2(x- dfrac{pi}{12})right]=sin (2x- dfrac{pi}{6}); By examining the options, it's not hard to find: When x= dfrac{pi}{12}, sin left(2times dfrac{pi}{12}- dfrac{pi}{6}right)=0; thereforeleft( dfrac{pi}{12},0right) is one of the centers of symmetry of the function. Therefore, the correct choice is boxed{D}. This problem is solved by determining the equation of the function after the translation and then identifying one of the centers of symmetry through the options. This basic question tests the understanding of the translation of trigonometric function graphs and the concept of centers of symmetry, focusing on computational ability and logical reasoning, a common question type.