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\centerline{MATH 392 -- Geometry Through History}
\centerline{Group Discussion -- One testimony about the impact of Euclid, January 29}
\vskip 10pt
In his brief biography of the philosopher Thomas Hobbes from about 1690 CE, John Aubrey wrote:
\vskip 10pt
\emph{He [that is, Hobbes] was . . . 40 yeares old before he looked on geometry; 
which happened accidentally.  Being in a gentleman's library in . . . , Euclid's 
{\it Elements} lay open, and `twas the 47 El. libri I  [that is, the 47th proposition 
in Book I].  He read the proposition. `By G--,' (he would now and then sweare an emphaticall 
Oath by way of emphasis) sayd he, `this is impossible!'  So he reads the demonstration of 
it, which referred him back to such a proposition; which proposition he read.  That
referred him back to another, which he also read.  \emph{Et sic deinceps} that at last
he was demonstratively convinced of that trueth.  This made him in love with geometry.}
\vskip 10pt 
\noindent
Today, we want to recreate something like the experience that Aubrey is referring to
by reading through the proof that Euclid gives for his Proposition 47 in Book I of 
the {\it Elements} and seeing how the results laid down previously prepare for this
surprising and elegant proof.   Here's the statement and the proof as Euclid gives it.
(Notes:  I have written this using some modern notation for quantities such as angles.
The translation from the original Greek is based on the standard version by T.L. Heath,
with some slight ``tweaks'' by J.L. to show the literal
wording more closely in a few instances):

\vskip 10pt 
\noindent
{\bf Proposition 47.}  In right-angled triangles the square on the side opposite 
the right angle is equal to the squares on the sides containing the right angle.

\vskip 10pt
\noindent
{\it Questions, part 1:  What is the usual name we give to this result and how do we usually
state it?  Euclid's point of view is different because he essentially does not have 
the ``infrastructure'' of algebra--it had literally
\emph{not yet been invented in his time}(!)--and he wants to state everything in a geometric form.  What sort of equality is this asserting (i.e. what two quantities are equal)?  Note:  Some of this
may only become clear as you work through the rest of the questions below.  If you don't see it now, 
come back later(!)}

\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{EuclidI47.jpg}
\caption{Figure for Proposition 47, Book I}
\end{center}
\end{figure}

\vskip 10pt 
\noindent
Note:  Refer to the figure at the top of the next page for all notation.
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\noindent
{\bf Proof:}  
Let $ABC$ be a right-angled triangle having the angle $\angle BAC$ right.

I say that the square on $BC$ is equal to the squares on $BA$ and $AC$.

Let the square $BDEC$ on the side $BC$ have been described, 
and [similarly] the squares $GB$ and $HC$ on $BA$ and $AC$ 
(Proposition 46).

\vskip 10pt
\noindent
{\it Questions, part 2: Look at the list of propositions in Book I given in McCleary's
book, pages 24-26.  What does Proposition 46 say?  Can you guess how that construction 
works and which of the earlier Propositions it relies on?}

\vskip 10pt
Let the line $AL$ through $A$ have been constructed parallel to either $BD$ or $CE$, 
and let $AD$ and $FC$ have been joined (Proposition 31, Postulate 1).

\vskip 10pt
\noindent
{\it Questions, part 3: Again look at the list of propositions in Book I given in McCleary's
book, pages 24-26 and the discussion of the postulates in the preceding section of the text.  
What does Proposition 31 say?  Do you know a way to construct that parallel line?
What does Postulate 1 say? }

\vskip 10pt
Since each of the angles $BAC$ and $BAG$ is right, it follows that with a straight line $BA$, 
and at the point $A$ on it, the two straight lines $AC$ and $AG$ not lying on the same 
side make the adjacent angles equal to two right angles, therefore $CA$ is in a straight
line with $AG$.  For the same reason $BA$ is also in a straight line with $AH$ (Proposition 14).

\vskip 10pt
\noindent
{\it Questions, part 4: What does Proposition 14 say?  Note:  Euclid doesn't use any
numerical measure for angles so he needs to work a bit harder.  
How could we derive the statement in Proposition 14 more simply?}

\vskip 10pt
Since the angle $\angle DBC$ equals the angle $\angle FBA$, for each is right, let the angle 
$\angle ABC$ have been added to each, therefore the whole angle $\angle DBA$ equals the whole 
angle $\angle FBC$ (Postulate 4, Common Notion 2).

\vskip 10pt
\noindent
{\it Questions, part 5: What do Postulate 4 and Common Notion 2 say?}

\vskip 10pt
Since $DB$ equals $BC$, and $FB$ equals $BA$, the two sides $AB$ and $BD$ equal the two sides $FB$ and $BC$ respectively, and the angle $\angle ABD$ equals the angle $\angle FBC$, therefore the base $AD$ equals the base $FC$, and the triangle $\Delta ABD$ equals the triangle $\Delta FBC$
(Proposition 4).

\vskip 10pt
\noindent
{\it Questions, part 6: What does Proposition 4 say?  What is our usual name for that fact?
What does it mean here to say the triangle $\Delta ABD$ equals the triangle $\Delta FBC$?
They're clearly not exactly the same triangle, so what kind of relation is this equality?}

\vskip 10pt
Now the parallelogram $BL$ is double the triangle $\Delta ABD$, for they have the same base $BD$ and are in the same parallels $BD$ and $AL$. And the square $GB$ is double the triangle $FBC$, for they again have the same base $FB$ and are in the same parallels $FB$ and $GC$ (Proposition 41).

\vskip 10pt
\noindent
{\it Questions, part 7: What does Proposition 41 say?  How would we usually say this sort
of thing using standard facts from elementary geometry today?  And to what does the ``double''
refer here?  What is being doubled?}

\vskip 10pt
Therefore the parallelogram $BL$ also equals the square $GB$ (Common Notion 1).  

\vskip 10pt
\noindent
{\it Questions, part 8: How is the use of ``equals'' here different from 
the use of the same word before part 6 of the questions above?  Explain the two different senses
of that word.}

\vskip 10pt
Similarly, if $AE$ and $BK$ are joined, it will be shown that the parallelogram $CL$ is
equal to the square $HC$. Therefore the whole square $BDEC$ equals the sum of the two squares $GB$ and $HC$ (Common Notion 2).

And the square $BDEC$ is described on $BC$, and the squares $GB$ and $HC$ on $BA$ and $AC$.

Therefore the square on $BC$ equals the sum of the squares on $BA$ and $AC$.

Therefore in right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle, which is what it was necessary to show.

\vskip 10pt
\noindent
{\it Questions, part 9: Read over the Aubrey quotation 
about Hobbes's experience again.  Do you see now what he
was getting at? }   

\vskip 10pt
\noindent
What kind of thinker is likely to get really excited about this
sort of demonstration of the truth of a statement?  Some food for thought--Albert Einstein once said:
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\noindent
\emph{We reverence ancient Greece as the cradle of western science. Here for the first time 
the world witnessed the miracle of a logical system which proceeded from step to step 
with such precision that every single one of its propositions was absolutely indubitable---I 
refer to Euclid's geometry. This admirable triumph of reasoning gave the human intellect 
the necessary confidence in itself for its subsequent achievements. If Euclid failed to 
kindle your youthful enthusiasm, then you were not born to be a scientific thinker.}

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