1. As shown in the figure, in a square ABCD, diagonal lines AC and BD intersect at point O, E is a point on the edge of CD, and AE and BD intersect at point M to connect CM. Point f is a point on the side of CB, and AF intersects DB at point n to connect CN. (1) If the angle CME=30 degrees and the angle CNF=50 degrees, find the degree of the angle EAF. Here is the picture address:

2. As shown in the figure, in the right-angled trapezoidal ABCD, AD\\BC, angle ADC=90 degrees, L is the median vertical line of AD, which intersects with AD at point M, and a square ABFE is made with waist AB as the side, so that EP is perpendicular to L and P, which proves that the following is the picture address:

1. It is proved that an outer angle of a triangle is equal to the sum of two non-adjacent inner angles, so MAO=MCO. Because the outer angle = 30 and the angle Mao =15 can prove that there is the same angle oan = 25. After e, EO is perpendicular to AD. AQB is all equal to AEO2EP+AD=2CD, which can be converted into 2EP+2AM=2AQ. Because AQB is all equal to AEO, AO=AQ means AM+MO=AQ means AM+EP=AQ, and the rest should be enough.