What is the name of this theorem of Jakob Steiner's, and why is it true?

Question

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.

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?

Answer 1

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 $C$.

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

$$x^2+c^2=(a+b)^2$$

$$x^2+d^2=(e+f)^2$$

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

$$y^2+a^2=(e+f)^2$$

$$y^2+b^2=(c+d)^2$$

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

$$z^2+e^2=(c+d)^2$$

$$z^2+f^2=(a+b)^2$$

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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.

$$a^2+c^2+e^2=b^2+d^2+f^2$$

Answer 2

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

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

Answer 3

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: $$ CE^2 + FB^2 + AD^2 = EB^2 + FA^2 + DC^2 \\= (AB^2 + AC^2 + BC^2) - (AE^2 + CF^2 + BD^2) $$

Answer 4

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.

Answer 5

(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:

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

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

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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.)

Answer 6

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.

Answer 7

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.