Pythagoras alternative

In a right angle triangle such as AFE the square on the two sides adjacent to the right angle is equal to the square on the hypotenuse.

The truth can be seen visually by inspecting the above figures, but I have included an abbreviated proof below.

Consider square ABCD

Construct square BGEF with F on AB and G on BC

Construct square DHEI with I on DC and H on AD

Triangle AFE is a right angle triangle with ang. AFE a right angle 

AF = HE

Square of AF is the same as square of HE = IDHE

Square of EF is EFBG

Rectangles AFEH & IEGC are equal as AF = HE = IE = CG & HA = EF = EG = IC

Triangle   AEF = EHA  (Diagonal of rectangle creates 2 equal triangles)

Triangle   IEC = GCE  (Diagonal of rectangle creates 2 equal triangles)

But rect. AFEH = rect. CGEI so 

 Triangle  AEF = EHA = IEC  = GCE

Construct second square and show all triangles are equal.

Construct square A1B1C1D1 with A1B1 = AB

J is point on A1D1 such that    A1J = AH

M is point on A1B1 such that    A1M = AF

Join JM to form right ang. triangle A1MJ

Triangles A1MJ and AFE are equal (2 sides and inc. angle equal)

Similarly 

L is point on B1C1 such that C1L = AH

B1M = A1B1 - A1M = AB - AF = BF = EF (  Remember ABCD, EFBG,  A1B1D1C1 are all squares)

So as shown before triangles B1LM and AFE are equal.

Similarly triangles C1KL and D1JK also equal triangle AFE

Show MLKJ is a square

Consider triangle AFE, 

         ^FAE + ^AEF = 1 rt. ang.   (angles of triangle = 2rt. ang.)

         ^FAE + ^AEF = ^A1MJ + ^B1ML

                     = 1 rt. ang.

But 

^A1JM + ^B1ML + ^JML = 2 rt. ang.  (A1MB1 is a straight line)

So              ^JML =  rt. ang

Similarly ^MLK, ^LKJ, ^KZM  =  rt. ang

ML = LK = KJ = JM =  AE (All triangles equal) 

So MLKJ is a square (All angles rt. ang. and all sides equal)

Show Pythagoras Rule holds

Further     Sq. MLKJ =  Sq. A1B1C1D1- 4 triangles A1MJ

So             Sq AE = Sq. ABCD- 4 triangles AEF

But            Sq. ABCD- 4 triangles AEF = Sq. AF + Sq. EF

SO                   Sq. AE = Sq. AF + Sq. EF