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Problem Description Triangle \( ABC\) has a right angle at \( B\), \( AB = 1\), and \( BC = 2\). The bisector of \( \angle BAC\) meets \( \overline{BC}\) at \( D\). What is \( BD\)? [asy]unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,1), B=(0,0), C=(2,0); pair D=extension(A,bisectorpoint(B,A,C),B,C); pair[] ds={A,B,C,D}; dot(ds); draw(A--B--C--A--D); label("\(1\)",midpoint(A--B),W); label("\(B\)",B,SW); label("\(D\)",D,S); label("\(C\)",C,SE); label("\(A\)",A,NW); draw(rightanglemark(C,B,A,2));[/asy]$$ \textbf{(A)}\ \frac {\sqrt3 - 1}{2} \qquad \textbf{(B)}\ \frac {\sqrt5 - 1}{2} \qquad \textbf{(C)}\ \frac {\sqrt5 + 1}{2} \qquad \textbf{(D)}\ \frac {\sqrt6 + \sqrt2}{2}$$ \( \textbf{(E)}\ 2\sqrt3 - 1\)

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