Below, I'll show that the result can be shown using Rolle's Theorem.
Let $P$ be the point $(a,f(a))$, and let $Q$ be the point $(b,f(b))$. We're given that $M=(\alpha,\beta)$ is a point on line $PQ$ with $\alpha\notin[a,b]$.
For each $x\in[a,b]$, let $g(x)$ be the slope between $M$ and the point $(x,f(x))$. It follows that
$$g(x)=\frac{f(x)-\beta}{x-\alpha}$$
for all $x\in[a,b]$. From the fact that $\alpha\notin[a,b]$, we get that $g(x)$ is defined for all $x\in[a,b]$. It follows that the function $g:[a,b]\to\Bbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. From the fact that $M$ is on line $PQ$, we get that $g(a)=g(b)$. Note, that our function $g$ therefore satisfies the hypotheses of Rolle's Theorem. Hence $g'(t)=0$ for some $t\in(a,b)$.
Using the quotient rule, we get that
$$\begin{align*}
g'(x) &= \frac{(x-\alpha)f'(x)-(f(x)-\beta)}{(x-\alpha)^2} \\
&= \frac{f'(x)}{(x-\alpha)}-\frac{(f(x)-\beta)}{(x-\alpha)^2} \\
&= \frac{1}{x-\alpha}\cdot\left(f'(x)-\frac{f(x)-\beta}{x-\alpha}\right) \\
&= \frac{1}{x-\alpha}\cdot\left(f'(x)-g(x)\right) \\
&= \frac{f'(x)-g(x)}{x-\alpha}
\end{align*}$$
And so by Rolle's Theorem, there is a $t\in(a,b)$ with $g'(t)=0$. Hence
$$\frac{f'(t)-g(t)}{t-\alpha}=g'(t)=0.$$
It follows that $f'(t)-g(t)=0$. Hence $f'(t)=g(t)$.
If we let $R$ be the point $(t,f(t))$, then the slope of line $MR$ is $g(t)$. Also, the slope of the tangent line to the curve $y=f(x)$ at $R$ is $f'(t)$. It follows that $MR$ is the tangent line to the curve $y=f(x)$ at $R$.
Best Answer
The mean value theorem says that if you have two distinct $c,d$ such that $f(c)=c$ and $f(d)=d$, then there is an $a\in (c,d)$ such that $f'(a)=1$. Therefore there can be at most one such $c$.