# Problem #2261

 2261 The area of $\triangle$$EBD$ is one third of the area of $3-4-5$ $\triangle$$ABC$. Segment $DE$ is perpendicular to segment $AB$. What is $BD$? $[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3)); pair[] ps={A,B,C,D,E}; draw(A--B--C--cycle); draw(E--D); draw(rightanglemark(E,D,B)); dot(ps); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,S); label("E",E,NE); label("3",midpoint(A--C),NW); label("4",midpoint(C--B),NE); label("5",midpoint(A--B),SW); [/asy]$ $\textbf{(A)}\ \frac{4}{3} \qquad\textbf{(B)}\ \sqrt{5} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad\textbf{(E)}\ \frac{5}{2}$ This problem is copyrighted by the American Mathematics Competitions.
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