In \triangle A B C, A B=3, B C=4, C A=5. Circle \omega intersects \overline{A B} at E and B, \overline{B C} at B and D, and \overline{A C} at F and G. Given that E F=D F and \frac{D G}{E G}=\frac{3}{4}, length D E=\frac{a \sqrt{b}}{c}, where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. Find a+b+c.