\(\text{Let $B,$ $A,$ and $D$ be three consecutive vertices of a regular $18$-gon. A regular heptagon is constructed on $\overline{AB},$ with a vertex $C$ next to $A.$ Find $\angle BCD,$ in degrees. }\)
+9466 measure of an interior angle of a regular n-gon = ( n - 2 ) * 180° / n
m∠BAD = ( 18 - 2 ) * 180° / 18 = 160°
m∠BAC = ( 7 - 2 ) * 180° / 7 = (900/7)°
Now let's try to find m∠DAC
m∠DAC + m∠BAD + m∠BAC = 360°
So...
m∠DAC = 360° - m∠BAD - m∠BAC = 360° - 160° - (900/7)° = (500/7)°
Notice that △BAC is an isosceles triangle because AB = AC
And also △ACD is an isosceles triangle because AD = AC
The measure of each base angle of an isosceles triangle = ( 180° - measure of vertex angle ) / 2
m∠ACD = ( 180° - m∠DAC ) / 2 = ( 180° - (500/7)° ) / 2 = (380/7)°
m∠ACB = ( 180° - m∠BAC ) / 2 = ( 180° - (900/7)° ) / 2 = (180/7)°
Now we can find m∠BCD
m∠BCD = m∠ACB + m∠ACD = (380/7)° + (180/7)° = 80°
