Because AB=AD

Angle A=60 degrees

So triangle ADB is an equilateral triangle.

So AB=BD

Angle ABD= Angle ABE+ Angle DBE=60 degrees.

Because the line segment BE rotates counterclockwise around the point e by an angle α.

So the angle BEF=60 degrees

BE=EF

So the triangle BEF is an equilateral triangle.

So BE=BF

Angle EBF= Angle DBE+ Angle DBF=60 degrees.

So angel Abel = angel ·DBF

So triangle Abbe and triangle DBF are congruent (SAS)

So AE=DF

2. Solution: Because the angle D= 1/2 and the angle A+90 degrees.

Because the angle α= 60 degrees

So the angle D= 120 degrees.

So angle ADE+ angle DAB= 180 degrees.

So parallel AB

So angle FCM= angle MAB

Angle MFC= Angle MBA

So the triangle MFC is similar to the triangle MBA (AA).

So CF/AB=FM/BM

Because triangle AEB and triangle DFB are congruent (proved)

So the triangle of AFB = the triangle of DFB.

Because AE/DE= 1/2.

So the S triangle AEB/S triangle DEB= 1/2.

So s triangle DFB/S triangle DEB= 1/2.

Because the area of quadrilateral DEBF =S triangle DEB+S triangle DFB=9 times the root number 3/4.

So the s triangle AEB=E times the root number 3/4.

Because the angle of the S triangle AEB =1/2 * AE * AB * Sindab =1/2 * AE * AB * Sin60.

Because AD=AE+DE=AB (authentication)

Because AE/DE= 1/2.

So AE/AB= 1/3.

So: 3 times the root number 3/4= 1/2*AE*AB*sin60.

So AE= 1

AB=3

Because CF= 1/2

So CF/AB= 1/6.

So FM/MB= 1/6.

So BM/BF=6/7.

In triangle AEB, it is obtained by cosine theorem:

BE^2=AE^2+AB^2-2*AE*AB*cos60

So BE= root number 7

Because BE=BF (confirmed)

So BF= root number 7

So BM=6 times the root number 7/7.