设AD与BF交点为H,AH∥GF,AH/FG=BA/BG→AH=BA*FG/BG=a*b/(a+b)→DH=AD-AH=a-a*b/(a+b),S△DBF=S△DFH+S△DBH=1/2EF*DH+1/2AB*DH=1/2DH*(EF+AB)=1/2*[a-a*b/(a+b)]*(a+b)=1/2a^2当a=1,b=3/2时，S△DBF=1/2*(1)^2=1/2
