A geometry problem, come on. Tell me to do it. Come on. Increase the bounty.

1) Congruence. Because be=bf; Angle abc = angle abf, ba = bc, so the triangle is congruent.

2) Because the triangle is congruent, the angle bec=angle bfa=60 degrees, and be=bf, so the triangle bef is an isosceles right triangle, so the angle afe=60-45=15 degrees.

1) Congruence. Proof: Because ABCD is square.

So ab=bc

Angle ABF = Angle EBC = 90 degrees.

Because bf=be

So the triangle abf and the triangle cbe congruence (sas) 2) because the triangle abf and the triangle cbe congruence (proven) so the angle afb = the angle bec

Because the angle bec = 60 degrees.

So the angle AFB = 60 degrees.

Because angular AFE + angular BFE = 60 degrees.

Because the angle EBF = 90 degrees.

Because bf=be

So the triangle bef is an isosceles right triangle.

So angular BFE = 45 degrees.

So the angle AFE=15 degrees.

Because the quadrilateral abcd is a square, so ab=bc, and the angle abc=90 degrees, because in the triangle abf and the triangle cbe, ab=cd, the angle, abc=angle, abf=be, so the triangle abf is all equal to the triangle cbe (2), because be=bf, the angle bef angle bfe=45 degrees, so the angle aef 180 degrees, 45 degrees, 135 degrees (mobile phone segmentation incompetence).

1) is congruent (2) the angle bec is equal to 60 degrees.

bf = be angle bce is equal to 30 degrees angle fab is equal to 30 degrees.

Square abcd bf = be angle fba = 90 degrees.

BA = BC angle feb = angle BFE = 45 degrees.

FBA= EBC=90 degrees, FEB= AFE+ FAB, triangle ABF and triangle CBE, congruence (corner edges) 45=x+30, AFE=15

1) Two triangles are congruent, because be=bf, cbe= abf=90 degrees, cb=ab (corner edges), so the two triangles are congruent.

2) From (1) AFB= CEB=60, and be=bf, so bef= bfe=45, so afe= afb- bfe=60-45=15

Congruence, for the following reasons, because the quadrilateral abcd is a square, so ab cb, the angle abc angle abf, there is because eb fb, so they are congruent. The second question is, because the angle bec is 60 degrees, therefore, the angle afb is equal to 60 degrees, and there is because, fb eb, ab is perpendicular to fc, so the angle efb angle feb is 45 degrees, so the angle afe is equal to 15 degrees. Seek satisfaction,

As shown in the figure, as p r1 oa, pq1 ob, and the vertical foot is r1 and q1, it is easy to prove that rt pr1r congruent rt pq1qpr=pq

So the conclusion is that when rotating, PR is equal to PQ.

I don't know what level of education you are, I'll see how I should tell you.

Connect EF, DM, and the angle FDE=FED=FMD, so the triangle FDM is similar to FCD, so that fm*fc=fd 2=FA2 can be obtained, and the triangle FMA is similar to FAC, so the angle FMA=FAC=FAB, so the angle MEC=DFM=FAM+FMA=FAB+FAM=BAM, and the angle MEC=BAM, so MEBA is a four-point circle, and the angle AMB=AEB=90 degrees is accompanied by the fiber.

Hope it helps, some small details have been omitted by Vertical Burn. If you don't understand it, you can ask stupidly.