What is the name of this result about isosceles triangles?

Question

$\Delta ABC$ is isosceles with $AB=AC=p$. $D$ is a point on $BC$ where $AD=q$, $BD=u$ and $CD=v$.

Then the following holds.

$$p^2=q^2+uv$$

I would like to know whether this result has a name.

Answer 1

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.

Answer 2

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!

Answer 3

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.

Answer 4

This result (with several proofs) appears in a note by L. Hoehn in The Mathematical Gazette (March 2000 pp 71-73). The author suggests, plausibly, that it must have been found many times, but notes that "it doesn't seem to appear in the mathematical literature". He does not refer to or propose any name for the result.

Answer 5

It can be seen as a very straight-forward corollary of the intersecting chords theorem. If $c$ is a circle with center $A$ and passing through $B$ and $C$, $c$ has radius $p$. Then $BC$ is a chord of $c$ and line $AD$ intersect $c$ forming a diameter, say at points $E$ and $F$ (suppose $E$ is the point closer to $D$). Then $ED=p-q$ and $FD=p+q$. Your result follows from the intersecting chord theorem: $ED\cdot FD=BD\cdot CD$.

Answer 6

Let $M$ be the midpoint of $BC$. Let $MD = d$. It is easy to see that $v - u = 2d$ and so $v = u + 2d$. By Pythagoras $$p^2- q^2 = MC^2 - MD^2 = (MC + MD)(MC - MD) = [(u + v)/2 + d][(u + v)/2 - d] = [(2u + 2d)/2 + d] [(2u + 2d)/2 - d] = (u + 2d)u = vu.$$

Hence $p^2 = q^2 + uv$.