Expert: Paul Klarreich - 4/24/2009

Question
Proof #8


Playing with the applet that demonstrates the Euclid's proof
(#7), I have discovered another one which, although ugly,
serves the purpose nonetheless.

Thus starting with the triangle 1 we add three more in the
way suggested in proof #7: similar and similarly described
triangles 2, 3, and 4. Deriving a couple of ratios as was
done in proof #6 we arrive at the side lengths as depicted
on the diagram. Now, it's possible to look at the final
shape in two ways:

as a union of the rectangle (1+3+4) and the triangle 2,
or
as a union of the rectangle (1+2) and two triangles 3 and
4.
Equating areas leads to

    ab/c · (a²+b²)/c + ab/2 = ab + (ab/c · a²/c
+ ab/c · b²/c)/2
Simplifying we get

    ab/c · (a²+b²)/c/2 = ab/2, or (a²+b²)/c² = 1
Remark

In hindsight, there is a simpler proof. Look at the
rectangle (1+3+4). Its long side is, on one hand, plain
c, while, on the other hand, it's a²/c+b²/c and we again
have the same identity.

FOr the diagram, please go to http://www.cut-the-
knot.org/pythagoras/index.shtml and scroll down toe the
proof 8.

Thankyou in advance.
Ps. it would b great is u could explain how the triangles
are similar, and if u can tell me the property that allows u
to do what u are doing in the problem so i know.

so far i simplified the equation but can't get a squared +
b squared = c squared.  
thx again.  

Answer
Questioner:   rohan
Country:  United States
Category:  Advanced Math
Private:  No
 
Subject:  Math Proof--Tough
Question:  Proof #8


FOr the diagram, please go to
http://www.cut-the-knot.org/pythagoras/index.shtml

and scroll down to the proof 8.

Thankyou in advance.

Ps. it would b great is u could explain how the triangles
are similar, and if u can tell me the property that allows u
to do what u are doing in the problem so i know.

so far i simplified the equation but can't get a squared +
b squared = c squared.  
thx again.
===============================
Hi, Rohan,

It took a while to figure this out (maybe that's what the author intended) but I think it goes like this:

The right triangles (2,3,4) are constructed so that each has the hypotenuse on one of a,b,c, and having the angles  theta and phi to match those in triangle (1).

[Yes, good labeling is the key to understanding, or at least to getting it right.]

Now match up x,y,z, with the angles in each triangle.  Note that phi + theta = 90, so x-z is a straight line and we do in fact have a rectangle.

In triangles (1) and (4), write this proportion (you check it out)

x     b
--- = ---
y     a

In ((3) and (1):
z     a
--- = ---
y     b

Now add those equations:

x + z    b     a
----- = --- + ---
 y      a     b

Note that x+z = c  and combine:

 c     b^2 + a^2
----- = ---------
 y        ab

      b^2 + a^2
c = y(-----------)   [EQ I]
         ab


Now in (1) (4) again:
y     a
--- = ---
b     c

    ab
y = ----
    c

Now put that back into [EQ I]:

   ab  b^2 + a^2
c = --(-----------)
   c      ab

And you can do the rest.