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}$
Solution
Two triangles sharing an angle have areas in the ratio of the products of the sides enclosing that angle (area $=\frac12ab\sin C$). We have $BC=9$, $DE=12$, $AF=15$, so $[ABF]:[ACE]=(3\cdot 15):(12\cdot 20)=3:16$, $[CDB]:[ACE]=(4\cdot 9):(12\cdot 16)=3:16$, $[DEF]:[ACE]=(12\cdot5):(16\cdot 20)=3:16$.
In right triangle $ACE$ (right angle at $C$), $AC=9$, $CE=12$, $EA=15$. Points $B,D,F$ lie on $AC,CE,EA$ respectively with $AB=3$, $CD=4$, $EF=5$. 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}$ 
