In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$  Find the ratio of $\angle ACB$ to $\angle APQ.$  Enter your answer as a fraction.
Let $x = \angle QBP = \angle QPB.$

[asy]
unitsize(6 cm);

pair A, B, C, P, Q;

A = (0,0);
B = dir(260);
C = dir(280);
P = extension(B, B + dir(70), A, C);
Q = extension(C, C + dir(130), A, B);

draw(A--B--C--cycle);
draw(Q--P--B);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$P$", P, NE);
label("$Q$", Q, W);
[/asy]

Then $\angle BQP = 180^\circ - 2x,$ and $\angle PQA = 2x.$  Since triangle $APQ$ is isosceles, $\angle QAP = 2x.$  Then $\angle APQ = 180^\circ - 4x,$ so $\angle QPC = 4x.$  Since $\angle QPB = x,$ $\angle BPC = 3x.$

Also, since triangle $ABC$ is isosceles,
\[\angle ABC = \angle ACB = \frac{180^\circ - \angle BAC}{2} = 90^\circ - x.\]By the Law of Sines on triangle $BCP,$
\[\frac{BC}{BP} = \frac{\sin 3x}{\sin (90^\circ - x)} = \frac{\sin 3x}{\cos x}.\]By the Law of Sines on triangle $PQB,$
\[\frac{PQ}{BP} = \frac{\sin x}{\sin 2x} = \frac{\sin x}{2 \sin x \cos x} = \frac{1}{2 \cos x}.\]Since $BC = PQ,$ $\frac{\sin 3x}{\cos x} = \frac{1}{2 \cos x},$ so
\[\sin 3x = \frac{1}{2}.\]Since $\angle APQ = 180^\circ - 4x,$ $x < \frac{180^\circ}{4} = 45^\circ,$ so $3x < 135^\circ.$  Therefore, $3x = 30^\circ,$ so $x = 10^\circ.$

Then $\angle ACB = 90^\circ - x = 80^\circ$ and $\angle APQ = 140^\circ,$ and the ratio we seek is $\frac{80}{140} = \boxed{\frac{4}{7}}.$