Is this a New Proof for Pythagorean Theorem

geometrysolution-verification

(For all those that it may concern, this is not a duplicate of my previous post, But starts in a similar way.)

A triangle with side lengths a, b, c with a height(h) that intercepts the hypotenuse(c) at (x , y) such that it is split into two side lengths, c = m + n, we can find Pythagoras theorem using the area of a right triangle and the slope equations of the height and the hypotenuse.

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We begin by finding the coordinate (x,y) by using the concept of the area fourmula, ab = hc:

x – value:

$bx = hm$

$x= \frac{hm}{b}$

$x = \frac{am}{c}$

y- value:

$ay = hn$

$y = \frac{hn}{a}$

$y = \frac{bn}{c}$

which gives us:

$(\frac{am}{c}, \frac{bn}{c})$

We can also find the (x,y) coordinates using the slope equations of the height and hypotenuse.

Height's equation:

$y = \frac{b}{a}x$

Hypotenuse's equation:

$y = \frac{-a}{b}x + a$

Now we can determine the (x ,y )intercept.

x – value:

$\frac{b}{a}x = \frac{-a}{b}x + a$

$\frac{b}{a}x + \frac{a}{b}x = a$

$x(a^2 + b^2) = a^2b$

$x = \frac{a^2b}{a^2 + b^2}$

y – value:

$y = \frac{b}{a}(\frac{a^2b}{a^2 + b^2})$

$y = \frac{ab^2}{a^2 + b^2}$

Which gives us:

$(\frac{a^2b}{a^2 + b^2}
,\frac{ab^2}{a^2 + b^2})$

Using $(\frac{bm}{c},\frac{an}{c}) ,(\frac{ab^2}{a^2 + b^2} , \frac{a^2b}{a^2 + b^2})$ there are 2 equalities:

$\frac{bm}{c} = \frac{a^2b}{a^2 + b^2}$

$\frac{an}{c} = \frac{ab^2}{a^2 + b^2}$

Which after isolating and eliminating c becomes:

$a^2m = b^2n$

This gives us two expressions:

$\sqrt{\frac{n}{m}} = \frac{a}{b}$

$\sqrt{\frac{m}{n}} = \frac{b}{a}$

We substitute these into $c = m + n$:

$c = m + n$

$\frac{c}{\sqrt{m}{n}} = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}}$

$\frac{ab}{\sqrt{m}{n}}c = a^2 + b^2 $

Using $h$, this can be written as:

$1 = (\frac{\sqrt{m}{n}}{h})(\frac{a^2 + b^2}{c^2})$

or:

$1 = (\frac{h}{\sqrt{m}{n}})(\frac{c^2 }{a^2 + b^2})$

Note, this leaves us with only two possibilities, the fractions are either inverses, or the numerator and denominator are equal. We know $h = ab/c$ and $h < c$, so $h ≠ a^2 + b^2$, the denominator can also be split into $hc, c$ but we know $hc = ab$ so $ab ≠ a^2 + b^2$, and we know the side lengths $m,n$ are smaller than $a,b$ which means the numerator and denominator are equal in the case of:

$a^2 + b^2 = c^2$

Best Answer

As with your previous question, you have given a valid proof of Pythagoras... if I've followed your argument correctly. First, I'll condense it, if only for my own benefit. (I also swap round $a,\,b$, because traditionally these are respectively opposite $A,\,B$.)

With $C=O$, the hypotenuse $y=b(1-x/a)$ meets $y=ax/b$ at $x=ab^2/(a^2+b^2)$, a proportion $b^2/(a^2+b^2)$ of the way from $A$ to $B$, so $AH=b^2c/(a^2+b^2)$; $HB$ follows similarly. Equating two expressions for $\cos\theta$ (i.e. using similar triangles), $c=\frac{(a^2+b^2)h}{ab}$, which by area formulae is $(a^2+b^2)/c$.

Of the "standard" proofs I know, yours is most similar to this. But that proof doesn't even need area formulae: similar triangles give$$a^2=c\cdot BH,\,b^2=c\cdot HA\implies a^2+b^2=AB\cdot AB=c^2.$$@S.Dolan's answer gives similar time-saving tips, albeit still using area formulae.

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