The Pythagoras Theorem

My point of departure is the Pythagoras Theorem:
Everybody with secondary education knows the Pythagoras Theorem. But I bet over 90% of people don't know why it is true (i.e. they don't know how to prove it)! While proving the Pythagoras Theorem appears to be difficult, I present below a classical proof by Euclid (circa 330 - 260 BC) which requires no more than elementary Geometry known to every secondary student!

Please feel free to skip the rest if Geometric Proof doesn't appeal to you.
First, draw squares ABFG, BCDE and CAHK. Obviously, our task is to prove the area of square ABFG is the sum of the area of square BCDE plus the area of square CAHK. Usually people are stuck at this point. But what Euclid did was to draw a few lines as follows:

. draw CM parallel to AG, cutting AB at N and FG at M respectively
. join BH and CG

With Euclid's hints, the picture now becomes much clearer as in above figure, we can then proceed as follows:

In ΔABH and ΔAGC,
....................... AB = AG ......................... (sides of square ABFG)
.................. ∠BAH = ∠BAC + ∠CAH = ∠BAC + 90°
.................. ∠GAC = ∠GAB + ∠BAC = 90° + ∠BAC
............... ∴ ∠BAH = ∠GAC
...................... AH = AC .......................... (sides of square CAHK)
............... ∴ ΔABH Ξ ΔAGC ...................... (SAS)

....... Area of ΔABH = ½*AH*HK
........................... = ½* (AH*HK)
........................... = ½*(area of square CAHK)
....... Area of ΔAGC = ½*AG*GM
........................... = ½*(AG*GM)
........................... = ½*(area of rectangle AGMN)

Since ΔABH Ξ ΔAGC as proved above,
................ then area of ΔABH = area of ΔAGC
So, ½*(area of square CAHK) = ½*(area of rectangle AGMN)
......... ∴ area of square CAHK = area of rectangle AGMN ......... (1)

Similarly,
............. area of square BCDE = area of rectangle BFMN ......... (2)

Combining (1) and (2) together,
area of square CAHK + area of square BCDE = area of rect AGMN
................................................................... + area of rect BFMN
............................................................... = area of square ABFG

i.e. .............................................. a² + b² = c² .............. Q. E. D.