Proving the Pythagorean Theorem: A Traditional and a Modern Approach

Date: 01/27/98 at 04:14:12
From: Gaby Neusihin
Subject: Pitagoras

Can you please explain how I can prove the Pythagorean theorem?
If you can, please send me more than one answer.
                                 
Thanks a lot, 
Gaby

Date: 01/27/98 at 12:32:50
From: Doctor Rob
Subject: Re: Pitagoras

Start with the triangle ABC, where C is the right angle. Let the 
length of side AB be c, of side BC be a, and of sice AC be b.

The traditional proof goes like this. On each side of the triangle
construct a square whose side length is the length of the side of the
triangle: ABDE, BCFG, and CAHI. Drop a perpendicular from C to AB,
meeting it at K, and extended to meet DE at J. Draw lines CE and BH.  
Prove triangles ACE and ABH are congruent by S.A.S. 

Now show that the area of square ACIH is twice the area of triangle 
ABH, and the area of rectangle AEJK is twice the area of triangle ACE. 
Conclude that the area of square ACIH equals the area of rectangle 
AEJK. Similarly, draw lines CD and AG. Prove triangles BCD and ABG are 
congruent by S.A.S.  

Now show that the area of square BCFG is twice the area of triangle 
ABG, and the area of rectangle BDJK is twice the area of triangle BCD.  
Conclude that the area of square BCFG equals the area of rectangle 
BDJK.  

Now add your two conclusions: the area of square BCFG (a^2) plus the 
area of square ACIH (b^2) equals the area of the two rectangles AEJK 
plus BDJK (which together make up square ABDE, of area c^2).

A more modern approach is this. Extend CB past B b units to point Z.
Extend CA past A a units to point Y. Then CZ = CY = a+b. C, Y, and Z 
are three of the four corners of a square CYXZ of side a+b, so 
construct X, possibly by erecting perpendiculars to CZ at Z and to CY 
at Y. The point of intersection of the perpendiculars will be X. 

Now along YX mark point W so that YW = b and WX = a, and along ZX mark 
point V so that ZV = a and VX = b. Connect up AW, VW, and BV. Then 
triangles ABC, AWY, WXV, and VBZ are congruent right triangles, with 
side lengths a, b, and c. Now compute the area of the whole square 
CYXZ, which is (a+b)^2, and compute the area of the four triangles and 
the central square: 4*(a*b/2) + c^2. Set these equal, expand, and 
subtract 2*a*b from both sides.

Another: Assume a > b (if b > a, exchange the letters a and b, and 
the letters A and B throughout). Erect perpendiculars to line AB at 
A and B. Measure along these perpendiculars on the same side as point 
C c units, and mark points D and E. Connect line DE to form a square 
ABDE of side c and area c^2. Now extend BC beyond C to F so that 
BF = a, and mark on side AC point G so that AG = b. Connect GE. Draw 
DF and extend it to meet GE at point H. Now triangles ABC, BDF, DEH, 
and EAG are congruent right triangles with sides a, b, and c. Compute 
the area of the square ABDE in two ways: c^2 = 4*(a*b/2) + (a-b)^2.  
Expand and simplify.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/