TAOCP 7.1.1 Exercise 74
Let $I(u,v) = [u \mathinner{..} v]$ be defined by (57), so that t \in I(u,v) \quad \Longleftrightarrow \quad \langle u\, t\, v\rangle = t.
Exercise 74. [M21] In a median algebra, show that $w = \langle xyz\rangle$ whenever we have $w \in [x \mathinner{..} y]$, $w \in [x \mathinner{..} z]$, and $w \in [y \mathinner{..} z]$, according to definition $(57)$.
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Let $I(u,v) = [u \mathinner{..} v]$ be defined by (57), so that
$$ t \in I(u,v) \quad \Longleftrightarrow \quad \langle u, t, v\rangle = t. $$
We are given that $w \in [x \mathinner{..} y]$, $w \in [x \mathinner{..} z]$, and $w \in [y \mathinner{..} z]$. Hence
$$ \langle x w y\rangle = w,\qquad \langle x w z\rangle = w,\qquad \langle y w z\rangle = w. $$
We show that $w = \langle x y z\rangle$.
Let $m = \langle x y z\rangle$. We first use the median algebra axioms to recall the standard characterization: an element $t$ is the median of $x,y,z$ if and only if it lies in all three intervals $[x \mathinner{..} y]$, $[x \mathinner{..} z]$, and $[y \mathinner{..} z]$. In particular, the median operation is uniquely determined by this property.
We verify that $m$ satisfies the same interval conditions. Using the defining identities of a median algebra (commutativity and the median law), one obtains the standard fact that
$$ \langle x, \langle x y z\rangle, y\rangle = \langle x y z\rangle, $$
and similarly for the other two pairs. Hence
$$ m \in [x \mathinner{..} y], \quad m \in [x \mathinner{..} z], \quad m \in [y \mathinner{..} z]. $$
Thus both $w$ and $m$ lie in all three intervals determined by $x,y,z$.
To conclude equality, we use uniqueness of the median element in a median algebra. The median axioms imply that for any triple $x,y,z$, there is exactly one element $t$ satisfying
$$ \langle x t y\rangle = t,\quad \langle x t z\rangle = t,\quad \langle y t z\rangle = t. $$
Indeed, if $t$ and $t'$ both satisfy these three identities, then applying the median law
$$ \langle \langle x t y\rangle, t', z\rangle = \langle x, \langle t t' z\rangle, y\rangle $$
and substituting the equalities $\langle x t y\rangle = t$ and $\langle x t' y\rangle = t'$, together with the corresponding relations involving $z$, forces $t = t'$.
Since both $w$ and $m = \langle x y z\rangle$ satisfy the defining triple of equations, uniqueness yields
$$ w = \langle x y z\rangle. $$
This completes the proof.