Download Proposition 47 in Book I

The Elements, Book I – the
“destination” – Proposition 47
MONT 104Q – Mathematical
Journeys: Known to Unknown
The goal we have worked toward




Proposition 47. In a right triangle, the square on the
hypotenuse is equal to the sum of the squares on the two
other sides.
The “Theorem of Pythagoras,” but stated in terms of areas (not as
an algebraic identity!)
The proof Euclid gives has to rank as one of the masterpieces of
all of mathematics, although it is far from the simplest possible
proof.
The thing that is truly remarkable is the way the proof uses just
what has been developed so far in Book I of the Elements.
Euclid's proof, construction


Let the right triangle be ᐃABC with right angle at A
Construct the squares on the three sides (Proposition 46) and
draw a line through A parallel to BD (Proposition 31)
Euclid's proof, step 1


<BAG + <BAC = 2 right angles, so G,A,C are all on one line
(Proposition 14), and that line is parallel to FB (Proposition 28)
Consider ᐃFBG
Euclid's proof, step 2

Proposition 37 implies areas of ᐃFBG and ᐃFBC are equal
Euclid's proof, step 3

AB = FB and BC = BD since ABFG and BCED are squares

<ABD = <FBC since each is a right angle plus <ABC

Hence ᐃBFC and ᐃABD are congruent (Proposition 4 – SAS)
Euclid's proof, step 4


Proposition 37 again implies ᐃBDA and ᐃBDM have the same
area.
Hence square ABFG and rectangle BLKD have the same area
(twice the corresponding triangles – Proposition 41)
Euclid's proof, conclusion


A similar argument “on the other side” shows ACKH and CLME
have the same area
Therefore BCDE = ABFG + ACKH. QED