# 1979 AHSME Problems/Problem 12

## Problem 12 $[asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("A",A,W); label("B",B,NW); label("C",C,S);label("D",D,E);label("E",EE,NE);label("O",O,S);label("45^\circ",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]$

In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$. Point $A$ lies on the extension of $DC$ past $C$; point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle. If length $AB$ equals length $OD$, and the measure of $\measuredangle EOD$ is $45^\circ$, then the measure of $\measuredangle BAO$ is $\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$

## Solution

Solution by e_power_pi_times_i

Because $AB = OD$, triangles $ABO$ and $BOE$ are isosceles. Denote $\measuredangleBAO = \measuredangleAOB = \theta$ (Error compiling LaTeX. ! Undefined control sequence.). Then $\measuredangleABO = 180^\circ-2\theta$ (Error compiling LaTeX. ! Undefined control sequence.), and $\measuredangleEBO = \measuredangleOEB = 2\theta$ (Error compiling LaTeX. ! Undefined control sequence.), so $\measuredangleBOE = 180^\circ-4\theta$ (Error compiling LaTeX. ! Undefined control sequence.). Notice that $\measuredangleAOB + \measuredangleBOE + 45^\circ = 180^\circ$ (Error compiling LaTeX. ! Undefined control sequence.). Therefore $\theta+180-4\theta = 135^\circ$, and $\theta = \boxed{\textbf{(B) } 15^\circ}$.

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