Let $B,$ $A,$ and $D$ be three consecutive vertices of a regular $20$-gon. A regular heptagon is constructed on $\overline{AB},$ with a vertex $C$ next to $A.$ Find $\angle BCD,$ in degrees. [asy] unitsize(2 cm); pair A, B, C, D, O; A = dir(0); B = dir(360/20); D = dir(-360/20); O = extension(A, A + rotate(-450/7)(B - A), B, B + rotate(450/7)(A - B)); C = A + rotate(-900/7)(B - A); for(int i = 0; i <= 19; ++i) { draw(dir(360/20i)--dir(360/20(i + 1))); } for(int i = 0; i <= 5; ++i) { draw((O + rotate(-360/7i)(B - O))--(O + rotate(-360/7(i + 1))*(B - O))); } dot("$A$", A, W); dot("$B$", B, W); dot("$C$", C, S); dot("$D$", D, W); [/asy]

Question

Let $B,$ $A,$ and $D$ be three consecutive vertices of a regular $20$-gon. A regular heptagon is constructed on $\overline{AB},$ with a vertex $C$ next to $A.$ Find $\angle BCD,$ in degrees. [asy] unitsize(2 cm); pair A, B, C, D, O; A = dir(0); B = dir(360/20); D = dir(-360/20); O = extension(A, A + rotate(-450/7)(B - A), B, B + rotate(450/7)(A - B)); C = A + rotate(-900/7)(B - A); for(int i = 0; i <= 19; ++i) { draw(dir(360/20i)--dir(360/20(i + 1))); } for(int i = 0; i <= 5; ++i) { draw((O + rotate(-360/7i)(B - O))--(O + rotate(-360/7(i + 1))*(B - O))); } dot("$A$", A, W); dot("$B$", B, W); dot("$C$", C, S); dot("$D$", D, W); [/asy]

Answer 1

Note that $\angle ABC = \angle BCA$ in an equilateral triangle, so $AB = BC$. By symmetry, $BC = CD$. Therefore, $BC = CD = AB$, so $BCD$ is a triangle with base angles all equal to $\frac{180^\circ - \angle BCD}{2}$. Thus, \[180^\circ = \angle BCD + 2\left(180^\circ - \angle BCD\right) = 180^\circ + 360^\circ - 3\angle BCD.\]It follows that $\angle BCD = \boxed{080}$.

Answer 2

Let's denote the center of the $20$-gon as $O.$ Since the $20$-gon is regular, we can see that $\angle AOB$ is equal to $\dfrac{360^\circ}{20} = 18^\circ.$

[asy] unitsize(2 cm); pair A, B, C, D, O; A = dir(0); B = dir(360/20); D = dir(-360/20); O = extension(A, A + rotate(-450/7)*(B - A), B, B + rotate(450/7)*(A - B)); C = A + rotate(-900/7)*(B - A); for(int i = 0; i <= 19; ++i) { draw(dir(360/20*i)--dir(360/20*(i + 1))); } for(int i = 0; i <= 5; ++i) { draw((O + rotate(-360/7*i)*(B - O))--(O + rotate(-360/7*(i + 1))*(B - O))); } draw(A--B--C--cycle,gray(0.7)); dot("$A$", A, W); dot("$B$", B, W); dot("$C$", C, S); dot("$D$", D, W); dot("$O$", O, N); [/asy]

Also, note that $ABCD$ is cyclic since $\angle AOB + \angle BOD = 18^\circ + 18^\circ = 36^\circ = 180^\circ.$

Since $ABCD$ is a cyclic quadrilateral, we know that $\angle ACD = 180^\circ - \angle ABD.$ Since the $20$-gon is regular, $\angle ABD = \dfrac{180^\circ}{20} = 9^\circ.$

Hence, $\angle ACD = 180^\circ - 9^\circ = 171^\circ.$

Since $\angle ACD + \angle BCD = 180^\circ,$ we can solve for $\angle BCD:$

$\angle BCD = 180^\circ - \angle ACD = 180^\circ - 171^\circ = \boxed{9^\circ}.$