# Problem #2136

 2136 In the right triangle $\triangle ACE$, we have $AC=12$, $CE=16$, and $EA=20$. Points $B$, $D$, and $F$ are located on $AC$, $CE$, and $EA$, respectively, so that $AB=3$, $CD=4$, and $EF=5$. What is the ratio of the area of $\triangle DBF$ to that of $\triangle ACE$? $\mathrm{(A) \ } \frac{1}{4} \qquad \mathrm{(B) \ } \frac{9}{25} \qquad \mathrm{(C) \ } \frac{3}{8} \qquad \mathrm{(D) \ } \frac{11}{25} \qquad \mathrm{(E) \ } \frac{7}{16}$ $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label("A",A,N); label("B",B,W); label("C",C,SW); label("D",D,S); label("E",E,SE); label("F",F,NE); label("3",A--B,W); label("9",C--B,W); label("4",C--D,S); label("12",D--E,S); label("5",E--F,NE); label("15",F--A,NE); [/asy]$ This problem is copyrighted by the American Mathematics Competitions.
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