2011 AMC 10B Problems/Problem 17

Problem 17

In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$ ?

$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$

Solution

$[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=3; pair O=(0,0), E=(-3,0), B=(3,0); path outer=Circle(O,r); draw(outer); draw(E--B); pair[] ps={B,E,0}; dot(ps); label("$B$",B,NE); label("$E$",E,NW); label("$$",O,N); [/asy]$

We can let $\angle AEB$ be $4x$ and $\angle ABE$ be $5x$ because they are in the ratio $4 : 5$ . When an inscribed angle contains the diameter, the inscribed angle is a right angle. Therefore by triangle sum theorem, $4x+5x+90=180 \longrightarrow x=10$ and $\angle ABE = 50$ .

$\angle ABE = \angle BED$ because they are alternate interior angles and $\overline{AB} \parallel \overline{ED}$ . Opposite angles in a cyclic quadrilateral are supplementary, so $\angle BED + \angle BCD = 180$ . Use substitution to get $\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(E)} 130}$

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2011 AMC 10B Problems/Problem 17

Problem 17

Solution