Are there curves other than circles such that the line through two points on the curve is parallel the line tangent to the curve at midpoint? In a circle, if we pick any two distinct points $p_1$ and $p_2$  and draw a line passing through $p_1$ and $p_2$  that line is parallel to the line tangent to the circle at the midpoint of the arc with endpoints $p_1$ and $p_2$.
Are there curves other than cicles such that is true? What about parabolas?
 A: In my opinion, this is a great question with a clear interpretation. I don't have an answer yet, but I can at least explain how to formulate the question precisely, which may help others find a solution.
Let's suppose the curve is a regular smooth plane curve parametrized by $\mathbb{R}$. Then without loss of generality it may be parametrized by arc-length, i.e. we have a smooth function $\gamma : \mathbb{R} \to \mathbb{R}^2$ such that $\lVert \gamma'(t) \rVert = 1$ for all $t \in \mathbb{R}$.
The condition that the secant line be parallel to the tangent at the midpoint then says that there exists a function $c : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ such that
$$\gamma(b) - \gamma(a) = c(a,b) \gamma'\left(\frac{a+b}{2}\right)$$
for all $a,b \in \mathbb{R}$. Using the fact that $\gamma'$ is a unit vector, we have
$$c(a,b) = c(a,b) \left\lVert \gamma'\left(\frac{a+b}{2}\right) \right\rVert^2 = c(a,b) \gamma'\left(\frac{a+b}{2}\right) \cdot \gamma'\left(\frac{a+b}{2}\right) = (\gamma(b) - \gamma(a)) \cdot \gamma'\left(\frac{a+b}{2}\right).$$
We wish to know if this forces $\gamma$ to be a circle or a line, i.e. a curve of constant curvature. So the question can be posed as:
Let $\gamma : \mathbb{R} \to \mathbb{R}^2$ be a smooth function such that
$$\lVert \gamma'(a) \rVert = 1$$
and
$$\gamma(b) - \gamma(a) = \left((\gamma(b) - \gamma(a)) \cdot \gamma'\left(\frac{a+b}{2}\right)\right) \gamma'\left(\frac{a+b}{2}\right)$$
for all $a, b \in \mathbb{R}$. Must $t \mapsto \lVert \gamma''(t) \rVert : \mathbb{R} \to \mathbb{R}$ be constant?
A: Improved answer
Let $\gamma$ be immersive.
As $\gamma(s+h)-\gamma(s-h)$ is parallel to $\gamma'(s)$, we have
$$\gamma(s+h)-\gamma(s-h)=\langle\gamma(s+h)-\gamma(s-h),\gamma'(s)/\|\gamma'(s)\|\rangle
\frac{\gamma'(s)}{\|\gamma'(s)\|}.$$
Now take the second derivative in respect to $h$ to get
$$\gamma''(s+h)-\gamma''(s-h)=\langle\gamma''(s+h)-\gamma''(s-h),\gamma'(s)/\|\gamma'(s)\|\rangle
\frac{\gamma'(s)}{\|\gamma'(s)\|}.$$
Divide by $2h$ and let $h\to0$:
$$\gamma'''(s)=\langle\gamma'''(s),\gamma'(s)/\|\gamma'(s)\|\rangle\gamma'(s)\frac{\gamma'(s)}{\|\gamma'(s)\|},$$
that is, $\gamma'''(s)$ is parallel to $\gamma'(s)$,
In this case
$$\left( \det(\gamma',\gamma'') \right)'=0,$$
hence
$$\kappa(s)=\frac{\det\bigl(\gamma'(s),\gamma''(s)\bigr)}{\|\gamma'(s)\|^3}
=\frac{\text{constant}}{\|\gamma'(s)\|^3},$$
that is, the curvature solely depends on the length of the velocity vector.  For an arc-length parametrisation it follows  that the curvature is constant.
As example would serve $\gamma(s)=(a\cdot e^s,b\cdot e^{-s})$ for constants $a$ and $b$.
Original answer
As $\gamma'(s)$ is a unit vector and $\gamma(s+h)-\gamma(s-h)$ is parallel to $\gamma'(s)$, we have
$$\gamma(s+h)-\gamma(s-h)=\langle\gamma(s+h)-\gamma(s-h),\gamma'(s)\rangle\gamma'(s),$$
as @diracdeltafunk remarked.
Now take the second derivative in respect to $h$ to get
$$
\gamma''(s+h)-\gamma''(s-h)=
\langle\gamma''(s+h)-\gamma''(s-h),\gamma'(s)\rangle\gamma'(s).
$$
Divide by $2h$ and let $h\to0$:
$$\gamma'''(s)=\langle\gamma'''(s),\gamma'(s)\rangle\gamma'(s),$$
that is, $\gamma'''(s)$ is parallel to $\gamma'(s)$, hence the derivative of $\|\gamma''(s)\|$ vanishes:
$$
\begin{align}
\frac{d}{ds}\|\gamma''(s)\|^2&=2\langle\gamma''(s),\gamma'''(s)\rangle\\
&=2\langle\gamma''(s),\langle\gamma'''(s),\gamma'(s)\rangle\gamma'(s)\rangle\\
&=2\langle\gamma'''(s),\gamma'(s)\rangle\langle\gamma''(s),\gamma'(s)\rangle\\
&=\langle\gamma'''(s),\gamma'(s)\rangle \frac{d}{ds} \lVert \gamma'(s) \rVert^2\\
&=0.
\end{align}
$$
