# 1977 AHSME Problems/Problem 9

## Problem 9 $[asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G); draw(B--E--C--A); label("A", A, NE); label("B", B, NW); label("C", C, SW); label("D", D, SE); label("E", E, E); //Credit to MSTang for the diagram [/asy]$

In the adjoining figure $\measuredangle E=40^\circ$ and arc $AB$, arc $BC$, and arc $CD$ all have equal length. Find the measure of $\measuredangle ACD$. $\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }\left(\frac{45}{2}\right)^\circ\qquad \textbf{(E) }30^\circ$

## Solution

Solution by e_power_pi_times_i

If arcs $AB$, $BC$, and $CD$ are congruent, then $\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta$. Because $ABCD$ is cyclic, $\measuredangle CAD = \measuredangle CBD = \theta$, and $\measuredangle ADB = \measuredangle ACB = \theta$. Then, $\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ$. $\theta = 55^\circ$. $\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{\textbf{(B) }15^\circ}$.