# 2011 AMC 10B Problems/Problem 17

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## Problem

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 $BCD$? $[asy] unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=3; pair A=(-3cos(80),-3sin(80)); pair D=(3cos(80),3sin(80)), C=(-3cos(80),3sin(80)); pair O=(0,0), E=(-3,0), B=(3,0); path outer=Circle(O,r); draw(outer); draw(E--B); draw(E--A); draw(B--A); draw(E--D); draw(C--D); draw(B--C); pair[] ps={A,B,C,D,E,O}; dot(ps); label("A",A,N); label("B",B,NE); label("C",C,S); label("D",D,S); label("E",E,NW); label("",O,N); [/asy]$ $\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$

## Solution 1

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 $\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(C)} 130}$

## Note:

We could also tell that quadrilateral $BEDC$ is an isosceles trapezoid because for $\overline{EB}$ and $\overline{DC}$ to be parallel, the line going through the center of the circle and perpendicular to $\overline{DC}$ must fall through the center of $\overline{DC}$.

## Solution 2

Note $\angle ABE = \angle BED=50$ as before. The sum of the interior angles for quadrilateral $EBCD$ is $360$. Denote the center of the circle as $P$. $\angle PDE = \angle PED = 50$. Denote $\angle PDC = \angle PCD = x$ and $\angle PBC = \angle PCB = y$. We wish to find $\angle BCD = x+y$. Our equation is $(\angle PDE +\angle PED)+(\angle PDC+\angle PCD)+(\angle PBC + \angle PCB) = 360 \longrightarrow 2(50) + 2x +2y = 360$. Our final equation becomes $2(x+y)+100 = 360$. After subtracting $100$ and dividing by $2$, our answer becomes $x+y=\boxed{\textbf{(C)} 130}$

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