In The Secrets of Triangles a remarkable theorem is attributed to Jakob Steiner.

Each side of a triangle is cut into two segments by an altitude. Build squares on each of those segments, and the alternating squares sum to each other. Steiner's altitude theorem

The book doesn't include a proof, and I'm not sure how to start.

Does this theorem have a name? How could one go about proving this beautiful relationship?

  • $\begingroup$ I haven't tried very hard to prove it yet. I should probably just play around with the Pythagorean Theorem to start. $\endgroup$
    – MBP
    Oct 23, 2017 at 23:13
  • 7
    $\begingroup$ By interesting coincidence, the Law of Cosines trigonograph that I posted this weekend may provide some insights. $\endgroup$
    – Blue
    Oct 23, 2017 at 23:20
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    $\begingroup$ That's really beautiful, Blue! $\endgroup$
    – MBP
    Oct 23, 2017 at 23:24
  • $\begingroup$ Yeah, I should have given this more of a go before posting. This isn't so bad at all. $\endgroup$
    – MBP
    Oct 23, 2017 at 23:25
  • $\begingroup$ It seems easy enough with carefully calculation: Determine $G$ and this determine all what remains. $\endgroup$
    – Piquito
    Oct 23, 2017 at 23:36

7 Answers 7


Label the squares' side lengths $a, b, c, d, e, f $ (clockwise from $A$). The claim is that $$a^2+c^2+e^2=b^2+d^2+f^2$$

Let $x$ be the altitude from $A$.

Let $y$ be the altitude from $B$.

Let $z$ be the altitude from $C$.

By the Pythagorean theorem applied to the two right triangles that include the altitude from $A$, we have:



By the Pythagorean theorem applied to the two right triangles that include the altitude from $B$, we have:



By the Pythagorean theorem applied to the two right triangles that include the altitude from $C$, we have:



Labeling the six Pythagorean equations above $(1)$ through $(6)$, we can add $(1)$, $(3)$, and $(5)$ to get:

$$ x^2+y^2+z^2 +a^2+c^2+e^2=(a+b)^2+ (c+d)^2 + (e+f)^2$$

Add $(2)$, $(4)$, and $(6)$:

$$ x^2+y^2+z^2 +b^2+d^2+f^2=(a+b)^2+ (c+d)^2 + (e+f)^2$$

Notice that the right sides of the above two equations are equal, so we may equate the left sides:

$$ x^2+y^2+z^2+a^2+c^2+e^2= x^2+y^2+z^2+b^2+d^2+f^2 $$

Now subtract $x^2+y^2+z^2$ from both sides, and we are done.


  • 6
    $\begingroup$ This is especially well-written. $\endgroup$
    – MBP
    Oct 24, 2017 at 0:04

Taking a cue from my Law of Cosines trigonograph, we have a straightforward arithmetic of areas:

enter image description here

$$\begin{align} \\ \\ \\ \color{red}{X_1} + \color{blue}{[\bullet\bullet\phantom{\bullet}]} &\quad=\quad \color{red}{X_2} + \color{green}{[\bullet\bullet\bullet]} &=\quad b c \cos A \\ \color{blue}{Y_1}\, + \color{green}{[\bullet\bullet\bullet]} &\quad=\quad \color{blue}{Y_2}\, + \color{red}{[\bullet\phantom{\bullet}\;\;\phantom{\bullet}]} &=\quad c a \cos B \\ \color{green}{Z_1} + \color{red}{[\bullet\phantom{\bullet}\;\;\phantom{\bullet}]} &\quad=\quad \color{green}{Z_2} + \color{blue}{[\bullet\bullet\phantom{\bullet}]} &=\quad a b \cos C \\ \hline \\ \color{red}{X_1} + \color{blue}{Y_1} + \color{green}{Z_1} &\quad=\quad \color{red}{X_2} + \color{blue}{Y_2} + \color{green}{Z_2} \end{align}$$

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    $\begingroup$ How long does it take you to make these beautiful geometry diagrams ? $\endgroup$
    – 6005
    Oct 24, 2017 at 17:55
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    $\begingroup$ @6005: Thanks for the compliment! :) Deciding what I actually want to draw usually takes the most time. In this case, that part went rather quickly because I had just been thinking about the Law of Cosines diagram in the past few days, and it was "obvious" that Steiner's result might follow fairly directly. In any case, once I have an idea, coaxing a diagram I like out of GeoGebra ---and then making tweaks in Photoshop--- isn't terribly difficult or time-consuming. I believe I posted my first pass at the answer above (which showed the diagram alone) within an hour of seeing the question $\endgroup$
    – Blue
    Oct 24, 2017 at 22:32

I am not sure it's all that complicated.

Note that by the Pythagoras theorem, all the following are true:

$CE^2 = AC^2 - AE^2$

$FB^2 = CB^2 - CF^2$

$AD^2 = AB^2 - BD^2$

$EB^2 = AB^2 - AE^2$

$FA^2 = AC^2 - CF^2$

$DC^2 = BC^2 - BD^2$

From the above, it is easy to see that: $$ CE^2 + FB^2 + AD^2 = EB^2 + FA^2 + DC^2 \\= (AB^2 + AC^2 + BC^2) - (AE^2 + CF^2 + BD^2) $$


Since nobody mentions the point $G$, here is another proof (although for this you have to use that the altitudes of a triangle always meet):

Consider the six small triangles involving $G$ (that is $AGF$, $GBF$ and so on) and invoke the Pythagorean theorem for each of them. Then for the grey squares we get

$$AF^2 + BE^2+ CD^2 = (AG^2 - FG^2) + (BG^2- EG^2) + (CG^2 - DG^2) $$

and for the red squares

$$ AD^2 + CE^2 + BF^2 = (AG^2 - DG^2) + (CG^2 - EG^2) + (BG^2-FG^2)$$ which are the same 6 terms, so both sums are equal.


(Community wiki.) As pertains to the claim that this was proved by Steiner, Viktor Blåsjö‏ points to Steiner's Werke, volume I, page 189 in a tweet accompanied by the following two images:

enter image description here

enter image description here

For completeness, here is the figure with the corresponding identity in $*$ illustrated geometrically:

enter image description here

where the summed area for the red and blue squares, respectively, are equal.

The above generalization, where the squares aren't determined specifically by altitudes (ie, by perpendiculars from the orthocenter) but rather by perpendiculars from an arbitrary point $P$, is known as Carnot's Theorem, not to be confused with Carnot's Theorem. Could this be one of those cases where the "wrong" name has been attached to a result? (Note that Steiner (1796–1863) and Carnot (1753–1823) were contemporaries.)

  • $\begingroup$ Can anyone explain how the diagram relates to the theorem? $\endgroup$
    – MBP
    Nov 1, 2017 at 1:09

HINT: Note that the right triangles $CBF$ and $CAF$ overlap on their leg $CF$. By the pythagorean theorem, $$AF^2+CF^2=CA^2$$ $$BF^2+CF^2=BC^2$$ and so $$CA^2-AF^2=BC^2-BF^2$$ or, by splitting up $CA=CD+DA$ and $BC=BE+EC$, $$\color{green}{(CD+DA)^2-AF^2=(BE+EC)^2-BF^2}$$ You should do the same thing with the other two altitudes to end up with a system of equations similar to the one I just derived, then try to manipulate the system to end up with the desired equality.


Take a triangle. Drop perpendiculars from corners to each side.

For each side bisected by a perpendicular, call the one clockwise black and counter clockwise red.

The perpendicular, the red half edge, and the triangle side touching the red half edge, form a right angle triangle. The triangle side part of this is clockwise from the corner the perpendicular drops from.

A similar result holds for the black half edge and the side of the triangle counterclockwise from the perpendicular corner.

So we get $$Red^2 + Perpendicular^2 = Side_{CW}^2$$ $$Black^2 + Perpendicular^2 = Side_{CCW}^2$$ Rearranging, we get $$Red^2 - Side_{CW}^2 = Black^2 - Side_{CCW}^2$$ or $$Red^2 = Black^2 + ( Side_{CW}^2 - Side_{CCW}^2 )$$ This is true for each perpendicular.

If we add up this truth for each perpendicular, the $( Side_{CW}^2 - Side_{CCW}^2 )$ term cancels out, because each original triangle side is clockwise from exactly one corner, and counter clockwise from exactly one corner.

This leaves

$$ \sum{Red^2} = \sum{Black^2} $$ as required.


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