(1) Because the angle FDE=60 degrees, the angle ADF+ the angle EDB= 120 degrees, and because ABC is an equilateral triangle, the angle B=60 degrees, so

Angle DEB+ angle BDE= 120 degrees, so angle ADF= angle bed, so triangular ADF is similar to triangular bed, let AF length be a,

AD=BD=b, FD=c, DE=d, so a/c=b/d, so a/b=c/d, because Angle A= Angle FDE, and a/b=c/d, so triangle ADF is similar to triangle DEF.

Problem (2) has not been solved. By the way, did you learn definition domain in ninth grade?