therefore

Because the diagonal angles are equal, ∠ AEB = ∠ NEM = 90. Similarly, it can be concluded that all four internal angles of a square are 90, so nmef is a square.

2. If the side length of a square is 1, then AE= 1. Therefore, the triangular aed is an isosceles triangle. Since aeb is a regular triangle, ∠ DAE = 30.

∠ Ade =∠ AED = 75, so∠ EDC =15.

3. As an auxiliary line AE, AE bisects the quadrilateral in the middle. Because the triangle ade is half of a regular triangle, the area of ade is 1*( 1/2)/2=0.25. So the area of the quadrilateral adeb is 0.25*2=0.5, and the area of the shaded part is 1-0.5=0.5.

I can't draw, so I don't know how to ask 1. EN is a constant, so I just need (EF

FN) Minimum is enough.

Make the symmetrical point e' of point e ABout ab, connect E'N, and ab crosses f, then point f is what you want. Because E'N=EN, let (EF

FN) That is (English)

FN) minimum value, as long as E'F and FN are on the same straight line. 2. Connect AC, AD, ab = ae, ∠B=∠E, BC=ED, ∴△ABC≌△AED, ∴AC=AD, ∴△ACM=MD are isosceles triangles, and the AM obtained by combining the three lines is also the center line of the CD.

E' is the point where E is symmetric about AB, that is to say, E' and E are symmetric about AB. The line segment between two points is the shortest, so the connection E'N can be (E'F).

FN) Minimum value. If point F is somewhere else, then E'FN becomes a triangle, and the sum of two sides of the triangle is greater than the third side, so it must be (E'F

FN) It's not as short as e 'n, and my explanation is very detailed. What else do I not understand?