What is the name of this result about isosceles triangles?

Stewart's theorem

Let $a$, $b$, and $c$ are the length of the sides of the triangles. Let $d$ be the length of the cevian to the side of length $a$.  If the cevian divides the side of length $a$ into two segments of length $m$ and $n$ with $m$ adjacent to $c$ and $n$ adjacent to $b$, then Stewart's theorem states that $b^2m +c^2n=a(d^2+mn)$

Note

Apollonius's theorem is a special case of this theorem.

Edit: I forgot some factors $\frac{1}{2}$.

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Derivation: $$h^2 = p^2 - \frac{1}{4}(u+v)^2\ ,$$ $$q^2 = h^2 + \frac{1}{4}(u-v)^2\ ,$$ where $h$ is the height perpendicular to $BC$. Therefore, $$p^2 = h^2 + \frac{1}{4}(u+v)^2 = q^2 - \frac{1}{4}(u-v)^2 + \frac{1}{4}(u+v)^2 = q^2 + uv\ .$$ I don't know if it has a name, but it is certainly a nice formula!

Besides Stewart's theorem, Ptolemy's theorem also gives this one as an immediate consequence: $ \def\len#1{\overline{#1}} \def\ang{\angle} \def\para{\parallel} \def\tri{\triangle} $

Let $E$ be the reflection of $A$ about the perpendicular bisector of $CD$. Then clearly $\len{CE} = q$ and $\len{DE} = p$. Also $DE \para AB$, so $\tri ABD \equiv \tri DEA$, and hence $\len{AE} = u$. Finally $ADCE$ is cyclic, so Ptolemy's theorem gives the desired result.