Why do waveforms that are symmetrical above and below their horizontal centerlines contain no even-numbered harmonics?
The complex Fourier coefficients of a $T$-periodic function $f(t)$ are given by

$$c_n=\frac{1}{T}\int_{0}^Tf(t)e^{-j2\pi nt/T}dt\tag{1}$$

with

$$f(t)=\sum_{n=-\infty}^{\infty}c_ne^{j2\pi nt/T}\tag{2}$$

The coefficients with even indices are

$$\begin{align}c_{2n}&amp;=\frac{1}{T}\int_{0}^Tf(t)e^{-j4\pi nt/T}dt\\&amp;=\frac{1}{T}\left[\int_{0}^{T/2}f(t)e^{-j4\pi nt/T}dt+\int_{T/2}^{T}f(t)e^{-j4\pi nt/T}dt\right]\\&amp;=\frac{1}{T}\left[\int_{0}^{T/2}f(t)e^{-j4\pi nt/T}dt+\int_{0}^{T/2}f(t+T/2)e^{-j4\pi nt/T}e^{-j2\pi n}dt\right]\\&amp;=\frac{1}{T}\int_{0}^{T/2}\left[f(t)+f(t+T/2)\right]e^{-j4\pi nt/T}dt\\&amp;=\frac12\frac{2}{T}\int_{0}^{T/2}\left[f(t)+f(t+T/2)\right]e^{-j2\pi nt/(T/2)}dt\\&amp;=\frac12 d_n\tag{3}\end{align}$$

where $d_n$ are the complex Fourier coefficients of the $T/2$-periodic function $g(t)=f(t)+f(t+T/2)$. These coefficients can only be zero if $g(t)=0$, i.e., if $f(t)=-f(t+T/2)$. This latter condition is exactly the symmetry condition you mention in your question. Consequently, the even Fourier coefficients are zero if (and only if) $f(t)$ satisfies

$$f(t)=-f(t+T/2)\tag{4}$$

Or, in other words, a $T$-periodic function $f(t)$ has only odd harmonics if it is $T/2$-antiperiodic.
I.e., if you shift the function by half its period and flip it across the horizontal axis, it must look the same as before.