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A math problem of "pointing training" in the second day of junior high school.
Solution:

∫ The quadrilateral ABCD is a diamond.

∴AB=AD,∠ABC=∠ADC

AE = AF = EF = AB

That is, AB=AE and AD=AF.

∴ ∠ABC=∠AEB,∠ADC=∠AFD

∠ABC=∠AEB=∠ADC=∠AFD

AB = AD

∴△ABE≌△ADF

∴∠BAE=∠DAF

∠ABE=( 180 degrees -∠BAE)/2

∫∠ABE+∠BAD = 180 degrees

∴∠ABE+∠BAD=( 180 degrees -∠BAE)/2+∠BAE+FAD+60 degrees =( 180 degrees-∠ BAE)/2+

∴90 degrees -2/3∠BAE= 120 degrees, ∠BAE=20 degrees

∴∠C=∠BAD=2∠BAE+60 degrees =20 degrees *2+60 degrees = 100 degrees.

A: The degree of angle C is 100 degrees.