I should apologize for my earlier rude and incorrect comment.
Here is a rather tedious proof that the median is a minimizer of $\min_x f(x)$ ($f$ defined below):
I need some notation: Let the set of points by $d_1,...,d_n$, let $\Omega = \{d_k\}_k$ (some points may be repeated) and define $f(x) = \sum_k |x-d_k|$. If $R$ is a relation on $\mathbb{R}$, let $I_R(x) = \{ k \,|\, x R d_k \}$ (eg, $I_=(x)$ is the set of indices such that $d_k = x$).
We note that $f$ is convex, and we have $f(x) = \sum_{k \in I_<(x)}d_k-x + \sum_{k \in I_>(x)}x-d_k$. Since $f$ is continuous, we see that $f$ is piecewise linear (affine), and for $x \notin \Omega$, $f'(x) = |I_>(x)|-|I_<(x)|$. Furthermore, since $f$ is convex, $f'(x)$ is non-decreasing on $\Omega^C$.
For $x < \min_k d_k$, we have $f'(x) = -n$, and for $x > \max_k d_k$, we have $f'(x) = +n$. Hence the problem $\min_x f(x)$ has a minimum on $[\min_k d_k, \max_k d_k]$.
Let $\phi_1,...,\phi_m$ be the non-decreasing sequence of values that $f'$ takes on $\Omega^C$, and let $J_1,...,J_m$ be the corresponding open intervals. (We have $\phi_1 = -n, \phi_m = +n$, of course.) There are two case to consider:
Case (1): For some $k$, we have $\phi_k = 0$, and for $x \in J_k$, $|I_>(x)|=|I_<(x)|$. It follows that $n$ is even, and since the median lies in $J_k$, we see that the median is a minimizer (in fact, $x$ is a minimizer iff $x \in \overline{J_k}$).
Case (2): For some $k$, we have $\phi_k <0, \phi_{k+1} >0 $. Let $\hat{x}$ be the point separating $J_k, J_{k+1}$ (hence $\hat{x} \in \Omega$). Then $\hat{x}$ is the minimizer of $f$. For $x \in J_k$, we have $f'(x) = |I_>(\hat{x})|-|I_=(\hat{x})|-|I_<(\hat{x})|$, and for $x \in J_{k+1}$, we have $f'(x) = |I_>(\hat{x})|+|I_=(\hat{x})|-|I_<(\hat{x})|$, or, in other words,
$$|I_=(\hat{x})| > \left| |I_>(\hat{x})|-|I_<(\hat{x})| \right| $$
To see why the median is $\hat{x}$, let $l=|I_>(\hat{x})|, p=|I_<(\hat{x})|$ and $e=|I_=(\hat{x})|$ ($l+e+p = n$, of course). The above shows that $e > |p-l|$. If $l+e+p$ is even, then let $k=\frac{1}{2}(e+p-l)>0$. Then we have $l+k = p+e-k$, from which it follows that the $l+k$th and $l+k+1$st (ordered) points are equal to $\hat{x}$, and hence $\hat{x}$ is the median.If $l+e+p$ is odd, then let $k=\frac{1}{2}(e+p-l-1)\geq 0$. Then the $l+k+1$st (ordered) point is equal to $\hat{x}$, and hence the median is equal to $\hat{x}$.
Alternative:
One may use the convex subdifferential to obtain the same characterization. A quick computation shows that $\partial f(x) = -\{|I_<(\hat{x})|\}+\{|I_>(\hat{x})|\}+[-|I_=(\hat{x})|,+|I_=(\hat{x})|]$, and $x$ is a minimizer iff $0 \in \partial f(x)$, or equivalently, $x$ is a minimizer iff $|I_=(\hat{x})| \geq \left| |I_>(\hat{x})|-|I_<(\hat{x})| \right| $. Then a little work (essentially the same as the cases above) shows that the median satisfies the latter characterization.
This is not something you can solve with geometric constructions for more than two points.
Suppose the $n$ points are $(x_i,y_i)$ and the target point is constrained to the $x$-axis, then the problem is equivalent to minimising the following function:
$$\sum_i\sqrt{(x-x_i)^2+y_i^2}$$
When considering the unconstrained problem, the set of points whose sum of distances to the $n$ points is constant is an $n$-ellipse, and the constrained problem is indirectly asking for that constant for which the $n$-ellipse is tangent to the $x$-axis. For two points, this shape is a normal ellipse, and the depicted reflection method works because the shortest distance between two points is a straight line (and the ellipse is of algebraic degree 2).
For three points, however, the 3-ellipse is of degree 8 – far too large to admit an exact solution in all cases.
I took your last diagram and assigned coordinates $(0,2),(2,-2),(3,3)$ to $A,B,C$ respectively. What is $P$ in your diagram – the $x$-intercept of the line between the Fermat point of $\triangle ABC$ and $B$ – lies at $(1.61509982\dots,0)$ where the sum of distances to $A,B,C$ is
$$7.911\mathbf64173\dots$$ This is quite good, but not optimal. The optimal point is $(1.59405306\dots,0)$ whose corresponding sum-distance is
$$7.911\mathbf42964\dots$$
That this differs only after the fourth decimal place shows how delicate this problem is, and the unchallenged superiority of numerical methods for tackling such problems.
Best Answer
This is Orthogonal Distance Regression. If you'd like to, there exists a scipy solution to it