13.(20 12? Lanzhou) as shown in the figure, in the quadrilateral ABCD, ∠ bad = 120, ∠ b = ∠ d = 90. Find a point m and n on BC and CD respectively to minimize the circumference of △AMN, then ∠ AMN+∠ ANM. )? a . 130 b . 120 c . 1 10d . 100

Examination Center: Axisymmetric-Shortest Path Problem. ?

Analysis: In order to minimize the perimeter of △AMN, that is, to make the three sides of the triangle on the same straight line by using the symmetry of points, and to make the symmetrical points A ′ and A″ of A about BC and ED, we can get ∠ aa ′ m+∠ a ″ = ∠ aha ′ = 60, and then get ∠ AMN.

Solution: Let A be the symmetry point A ′, A″ is about BC and ED, connecting A ″″, BC intersects M, and CD intersects N, then A ″″ is the minimum circumference of △AMN. Let A be the extension line of DA, AH, ∫≈eab = 120, ∴∴. And ∠ ma' a+∠ MAA' = ∠ AMN, ∠ NAD+∠ A "=

Comments: This topic mainly examines the solution of the shortest path problem in the plane, the nature of the outer angle of the triangle and the nature of the middle vertical line. According to the known knowledge, it is the key to solve the problem to get the position of m and n.