1。 Angle afp is the outer angle of triangle abf, so angle afp= angle baf+ angle abf, that is, angle afg+ angle gfc= angle baf+ angle abf, where angle abf= angle afg=90 degrees, so angle baf= angle cfg.

2 From the point of view of the problem, ae=fc, angle baf= angle cfg, and because the triangle bef is an isosceles right triangle, the angle bef is 45 degrees, and the angle aef= angle fcg= 135 degrees, the two triangles are congruent, according to asa.

3. If gm is perpendicular to bp when crossing the G point, it can be proved that the triangle abf is congruent with the triangle fmg, gm=eb= half the side length of the square, connecting e G, eg is parallel to ad and bp, while the regular triangle dae is congruent with the triangle fmg, and the angle ade is equal to the angle mfg. Because of parallelism, angle ade is equal to angle Deg, angle pfc is equal to angle egf, so angle deg is equal to angle fge, so ed is equal to fg.