[asy][asy]
import olympiad;
pair B = (0,0);
pair A = (0.5,1);
pair C = (1,0);
pair D = (0.5,0);
pair F = (0.25,0.5);
pair E = (0.8,0.4);
dot(A^^B^^C^^D^^E^^F);
draw(A--B--C--A);
draw(D--E--F--D);
draw(circumcircle(D,E,F));
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,S);
label("$E$",E,NE);
label("$F$",F,NW);
[/asy][/asy]
If the circumcircle of $\bigtriangleup FDE$ is tangent to $BC$. Then $\bigtriangleup BFD \sim \bigtriangleup DFE \sim \bigtriangleup CDE$. Thus $\frac{BD}{DE} = \frac{DF}{FE}$ and $\frac{CD}{DF} = \frac{DE}{FE}$. $$\Longrightarrow FE = \frac{DE \cdot DF}{CD} = \frac{DE \cdot DF}{DC}$$$$\Longrightarrow BD = DC$$