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### Proving the Pythagorean Theorem: A Traditional and a Modern Approach

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

Can you please explain how I can prove the Pythagorean theorem?

Thanks a lot,
Gaby
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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/
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Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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