I was reading Stewart's Theorem which states that

Given a triangle with side lengths $a$, $b$, $c$ and a cevian of length $d$ which divides $a$ into two segments $m$ and $n$ as shown in the figure

enter image description here

we must have $$b^2m+c^2n=a(d^2+mn)$$ which is often remembered using the funny mnemonic, "A man and his dad put a bomb in the sink" which gives $$man+dad=bmb+cnc$$

Now, this can be proved quite easily using The Law of Cosines on $\theta$ and $\theta^\prime$ as marked in the figure. You can find the proof here.

But, I was thinking if there is a more elementary geometric argument (without using trignometry and without using a lot of algebraic manipulation) to see this clearly in front of our eyes. I tried giving an attempt, but it seems extremely difficult using only elementary geometry since there is no trivial symmetry immediately observable in this diagram. Also, the terms involved in the formula, i.e., $b^2m$, $c^2n$, $ad^2$ and $amn$, are all of dimension $[L^3]$- so, it seems difficult to even draw areas on sides and geometrically manipulate them.

Any elegant geometric insight or an intuitive idea would be appreciated.

Note: as pointed out in the comments, and as you can see here as well, the Cosine Law can be very well proved using elementary geometry. While that is of course an answer, we all agree that it's cheating at some level. There's another answer as well, but this one's so much algebraic jargon, I would prefer trigonometry instead.

Also, Stewart's book, Some general theorems of considerable use in the higher parts of mathematics seems to contain a completely geometric argument. But the absense of diagrams in Google Scans combined with the length of this solution makes it very difficult to follow. I would also love to see a summarized version of this in more understandable way without loosing the geometric essence.

  • $\begingroup$ You may find the following helpful: math.stackexchange.com/questions/46616/… $\endgroup$
    – dodoturkoz
    Aug 23, 2021 at 17:29
  • $\begingroup$ @Sayan Dutta.- Do not forget that when a mathematical property has someone's name it is because that property is not obvious to the naked eye in any way. This does not prevent that it can be proved in other different ways but I do not think that in two lines. $\endgroup$
    – Piquito
    Aug 23, 2021 at 19:25
  • 5
    $\begingroup$ Well, Stewart's theorem follows from two applications of the Cosine theorem, and the cosine theorem follows from two applications of the Pythagorean theorem, so you can clearly prove Stewart's theorem with four applications of the Pythagorean theorem. $\cos\theta$ is just a convenient name for a ratio of (signed) lengths, so the boundary between "elementary" geometry and trigonometry is pretty much non-existent. $\endgroup$ Aug 23, 2021 at 21:18
  • $\begingroup$ @dodoturkoz That was of some help yes, but I couldn't understand the geometric arguments in the attached book. Please check the edit. $\endgroup$ Aug 24, 2021 at 5:55
  • $\begingroup$ @Piquito Well, maybe that's true for some theorems, but mostly not in case of elementary geometry. For example, consider the Pythagoras' Theorem and Euclid's beautiful proof of it. The congruency and area arguments make it clearly visible to the naked eye. $\endgroup$ Aug 24, 2021 at 5:57

4 Answers 4


enter image description here

$\triangle ABC$ is our reference triangle. Draw the circle centered at $A$ with radius $c$. Extend segments as shown in the figure. Using the power of a point at points $D$ and $C$, we get two equations:

$$c^2-x^2=m\cdot (n+y) \;\; \text{and} \; \; c^2-b^2=(m+n)\cdot y$$

Solving simultaneously (eliminating $y$) yields Stewart's Theorem. $\;\blacksquare$

What is meant by elementary geometry is rather controversial (power of a point $=$ similarity).

  • $\begingroup$ Can you please tell me which app/software you used to draw this diagram... $\endgroup$ Aug 25, 2021 at 20:14
  • $\begingroup$ Also, I don't know why I forgot to thank you for this solution. It's really brilliant (+1)! $\endgroup$ Aug 25, 2021 at 20:14
  • $\begingroup$ @SayanDutta Thank you! I also use GeoGebra 5 desktop version for producing the diagrams (the newer versions seem harder to navigate). Another frequently used software is Asymptote (it also produces very nice diagrams). If you want a quick head start you can read the last few bullet points of Evan Chen's article: web.evanchen.cc/handouts/Constructions/Constructions.pdf $\endgroup$
    – dodoturkoz
    Aug 26, 2021 at 3:47
  • $\begingroup$ Thanks, I'll check them out. $\endgroup$ Aug 26, 2021 at 4:29

Write $m:=a\mu$ and $n:=a\nu$. The highlighted similar triangles $\triangle A'DB$, $\triangle B''B'B$, $\triangle C''C'A$ in the figure (whose construction is detailed below) have corresponding sides in proportion, say, $\lambda:\mu$; we deduce $\triangle ABC\sim\triangle B''BA'$, so that $|A'B''|=b\lambda=|AC''|$, and we can write $|AA'|=|AC''|+|AB''|$.

enter image description here

Now, we simply apply the Intersecting Chords/Secants theorems thusly

$$ \begin{align} \color{green}{a\mu\cdot a\nu} = \overbrace{\color{green}{|AD|\,|A'D|}}^{\bigcirc ABC} &= |AD|\,(\,|AA'|-|AD|\,) \\[0.5em] &= |AD|\,(\,\color{blue}{|AC''|}+\color{red}{|AB''|}-|AD|\,) \\[0.5em] &=\overbrace{\color{blue}{|AC|\,|AC'|}}^{\bigcirc CC'C''}+\overbrace{\color{red}{|AB|\,|AB'|}}^{\bigcirc BB'B''} - |AD|^2 \\[0.5em] &=\color{blue}{b\cdot b\mu}+\color{red}{c\cdot c\nu} - d^2 \tag{$\star$} \end{align}$$

giving the equivalent of Stewart's Theorem. $\square$

The diagram is created by extending cevian $\overline{AD}$ to meet the circumcircle at $A'$. Then $B'$ and $C'$ divide $\overline{BA}$ and $\overline{AC}$ in the ratio $\mu:\nu$, and circumcircles $\bigcirc DBB'$ and $\bigcirc DCC'$ meet the cevian in points $B''$ and $C''$. A little angle-chasing establishes the similarity of the highlighted triangles.

Admittedly, that's quite a bit of setup —and perhaps there's an easier path to the key relation $|AA'|=|AB''|+|AC''|$— but I rather like how $(\star)$ falls right out.

  • 1
    $\begingroup$ This was brilliant (+1). I'm really pleased at the excellent answers this question is recieving! $\endgroup$ Aug 25, 2021 at 9:49
  • 2
    $\begingroup$ I wish more textbooks were following a similar style, presentation, and clarity. $\endgroup$
    – dodoturkoz
    Aug 25, 2021 at 12:14
  • $\begingroup$ Can you please tell me which app/software you used to draw this diagram... $\endgroup$ Aug 25, 2021 at 20:14
  • $\begingroup$ @SayanDutta: I use GeoGebra for my diagrams. $\endgroup$
    – Blue
    Aug 25, 2021 at 20:27
  • 1
    $\begingroup$ @Blue Okay, thanks. $\endgroup$ Aug 26, 2021 at 4:29

I have found a proof through Ptolemy's theorem and the intersecting chords theorem, but it is rather messy, sketched in the following diagram (beware: I accidentally swapped $m$ and $n$) enter image description here

Let us consider two spiral similarities bringing $ACD$ to $ABE$ and $ADB$ to $AFC$. Since we have supplementary angles at $D$, $ADBE$ and $AFCD$ are similar and cyclic. Their sides and diagonals are proportional to the sides and diagonals of the purple quadrilateral in the top right. The lengths of the diagonals are fixed by Ptolemy's theorem. Let us consider the intersection $K$ between $AD$ and the circumcircle of $ABC$: the quadrilateral $ABKC$ is similar to the quadrilateral with the orange diagonal, so $AK=\frac{c^2 m+b^2 n}{ad}$. On the other hand, by the intersecting chords theorem, $AK=d+\frac{mn}{d}$. This proves Stewart's theorem.

  • $\begingroup$ Nice one (+1)!! $\endgroup$ Aug 25, 2021 at 13:40
  • $\begingroup$ Can you please tell me which app/software you used to draw this diagram... $\endgroup$ Aug 25, 2021 at 20:14
  • $\begingroup$ I used Geogebra, too $\endgroup$ Aug 25, 2021 at 23:00
  • $\begingroup$ Okay, thanks. I'll check it out. $\endgroup$ Aug 26, 2021 at 4:31


This can be proved easily when triangle is right angled; suppose the foot of height is H, we have:

$$\triangle AHB\approx \triangle ACH$$

which results in:

$$ \frac bc=\frac c m\Rightarrow bm=c^\space\space\space\space\space\space\space(1)$$

$$\frac cd=\frac bn \Rightarrow cn=bd\space\space\space\space\space\space\space(2)$$

$$d^2=mn\space\space\space\space\space\space\space\space (3)$$

$bm=c^2\Rightarrow b^2m=(bc)d=2s\cdot c$

$cn=bd\Rightarrow c^2n=(bc)d=2s\cdot d$

$\Rightarrow bm^2+cn^2=2s(d+c)=d\cdot a (d+c)=a(d^2+c\cdot d)$

Multiplying (1) by (2) we obtain:

$bc(mn)=bc\cdot cd$

$\Rightarrow mn=cd$

Which gives:


This is true for all right angled triangles constructed when vertex A moves on a circle its diameter is BC.If we keep BC constant, the fact that this relation is correct for all types of triangles shows that any arbitrary transformation of a right angled triangle, keeping BC constant, to any arbitrary shape keeps the quantity the formula describe.

  • $\begingroup$ Can you please explain the last two statements.... I don't think I get it properly :( $\endgroup$ Aug 24, 2021 at 14:22
  • $\begingroup$ @SayanDutta. If you mean multiplying (1) by (2) it means both sides of relations are multiplied. $\endgroup$
    – sirous
    Aug 24, 2021 at 14:31
  • $\begingroup$ what I meant was "This true for all right angled triangle constructed when vertex A moves on a circle its diameter is BC. The fact that this relation is correct for any no right angled triangle shows that any arbitrary transformation of a right angled triangle preserves the quantity the formula indicates." $\endgroup$ Aug 24, 2021 at 14:33
  • $\begingroup$ @SayanDutta, I edited that part. I hope it is now clear. $\endgroup$
    – sirous
    Aug 24, 2021 at 14:44

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