The line segment in the euclidean metric I want to prove that for $a,b \in \mathbb{R}^2$ the set $S= \left\{x \in \mathbb{R}^2: d(a,b)=d(a,x)+d(x,b) \right\}$ coincides with the line segment connecting $a$ and $b$, where $d$ is the euclidean metric.
I was able to prove that, if $x$ lies in the line segment connecting $a$ and $b$, then $x$ also lies in $S$, assuming $x= \lambda a + (1-\lambda) b$ for some $0 \leq \lambda \leq 1$.
However, I was not able to prove that every point in $S$ lies in this line segment. It seems to be a little trickier. Can somebody help me?
Thank you!
 A: Take a point $x \in S$,suppose it does lie in the line connecting $a$ and $b$, then $x$, $a$, and $b$ form a triangle hence we will either have
$$d(a,b) > d(a,x) + d(x,b)$$
 or
$$d(a,x) > d(a,b) + d(x,b) \geq d(a,b)$$
or 
$$d(x,b) > d(a,b)+d(a,x) \geq d(a,b).$$
In each case, we can find a contradiction with the condition of $x \in S$.
Suppose $x$ doesn't lie in the convex hull of $a$ and $b$.
Then we either have $$d(a,x)=d(a,b)+d(b,x)$$
or $$d(b,x)=d(a,b)+d(a,x).$$
We can find a contradiction in either case.
For example, if $$d(a,x)=d(a,b)+d(b,x)$$
then $d(a,b)=d(a,b)+d(b,x)+d(b,x)$
Hence $x=b$ which contradicts with $x$ does lie in the convex hull of $a$ and $b$.
A: The following result is quite elementary, but the algebraic techniques will be reused.
Lemma 1: For all real numbers $a$ and $b$,
$\tag 1 | a + b | \le |a| + |b|$
Moreover, if both $a$ and $b$ are nonzero, then equality holds exactly when both the numbers have the same sign.
Proof
Use the $|c| = \sqrt {c^2}$ formulation for absolute value and after squaring both sides, (1) is deduced from algebra and elementary arithmetic laws.
Lemma 2: For all real numbers $x_1 \text{, } y_1$ and $ x_2 \text{, } y_2$
$\tag 2 \left(x_1^2 + y_1^2\right)\left(x_2^2 + y_2^2\right) - (x_1 x_2 + y_1 y_2)^2 = (x_1 y_2 - x_2 y_1)^2$
Proof: Equation (1) is an algebraic identity. QED
Using the Pythagorean distance norm for vectors, the triangle inequality emerges,
Proposition 3: For any two nonzero vectors $\vec v$ and $\vec w$ in $\mathbb{R}^2$,
$\tag 3 \|\vec v + \vec w \| \le \| \vec v \| + \| \vec w\|$
Moreover, equality occurs if and only if there exists a unique $\lambda \gt 0$ such that $\vec v = \lambda \vec w$.
Proof
To prove (3), set $\vec v = (x_1,y_1)^t$ and $\vec w = (x_2,y_2)^t$, and apply Lemma 2 with some algebraic technique.
For the second part, begin by assuming both sides of (3) are equal. Then, by Lemma 2, it must be true that
$\tag 4 x_1 y_2 - x_2 y_1 = 0$
Since both $\vec v$ and $\vec w$ are nonzero vectors, if say $x_1 = 0$, then $x_2 = 0$ and Lemma 1 can be used. So we can assume that the numbers  $x_1$, $y_1$ and $x_2$, $y_2$ are all nonzero. But then $y_1/y_2 = x_1/x_2$ and by setting $\lambda = y_1/y_2$, (4) will hold true. It can be easily checked that $\lambda$ must be positive for (3) to become an equality.
To show that (3) becomes an equality when $\vec v = \lambda \vec w$ for positive $\lambda$ is straightforward. QED
Since the distance between points is invariant under translations, the OP's question can be equivalently recast as follows,
Proposition 4: Let $\vec u$ be a nonzero vector in $\mathbb{R}^2$. Then the line segment connecting $\vec 0$ with $\vec u$ is equal to
$\tag 4 S=\{ \vec s \in \mathbb{R}^2 : \|\vec u\| = \|\vec s\|  + \| \vec u - \vec s\|   \}$
Proof
The line segment is equal to $\{ \alpha \vec u : 0 \le \alpha \le 1 \}$. Apply Proposition 3 and some more algebra to wrap up this proof. QED

Observe that the question was answered without using trigonometry or facts about the dot product. But the 'dot product algebra' is encountered in the above derivations, and is a natural way to 'discover' the definition of the inner product.
