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Pythagoras Theorem

Revision as of 15:43, 17 June 2019 by Colball (talk | contribs) (→‎Proof 2)

What is the Pythagorean Theorem?

What is the Pythagorean Theorem?

The Pythagoras Theorem is also referred to as the Pythagorean Theorem$.$ Pythagorean Theorem is used to find a side of any right triangle. It is $a^2+b^2=c^2$, where $a$ and $b$ are the legs of the triangle, and $c$ is the hypotenuse.

Why is it useful?

To find sides and angles of right triangles. Also, Trigonometry is pointless without it. If you know three angles of a triangle you can use the Pythagorean Theorem to find the sides or the area even if the angles are not right. It is probably the most famous Theorem in all of math!

Can we prove it?

Yes! The are hundreds of proves. I will just show you a few of them. Mathematicians even make a hobby of finding these proves. Even a US president made a published proof! Of, course this was a president in the 1800s because, well, presidents now are not really up to proving something like that. (You know, Trump and the others).

Proofs

Proof 1

We use $[ABC]$ to denote the area of triangle $ABC$.

Let $H$ be the perpendicular to side $AB$ from ${} C$.

[asy] pair A, B, C, H; A = (0, 0); B = (4, 3); C = (4, 0); H = foot(C, A, B); draw(A--B--C--cycle); draw(C--H); draw(rightanglemark(A, C, B)); draw(rightanglemark(C, H, B)); label("$A$", A, SSW); label("$B$", B, ENE); label("$C$", C, SE); label("$H$", H, NNW); [/asy]

Since $ABC, CBH, ACH$ are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,

$\frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2}$.

But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ∎

Proof 2

Consider a circle $\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\omega$. Let the line $AB$ meet $\omega$ at $Y$ and $X$, as shown in the diagram:

Pyth2.png

Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\omega$, we see

$AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ∎

Proof 3

$ABCD$ and $EFGH$ are squares.

[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label("$A$", A, NNW); label("$B$", B, ENE); label("$C$", C, ESE); label("$D$", D, SSW); draw(E--F--G--H--cycle); label("$E$", E, N); label("$F$", F,SE); label("$G$", G, S); label("$H$", H, W); label("a", A--B,N); label("a", B--F,SE); label("a", C--G,S); label("a", H--D,W); label("b", E--B,N); label("b", F--C,SE); label("b", G--D,S); label("b", A--H,W); label("c", E--H,NW); label("c", E--F); label("c", F--G,SE); label("c", G--H,SW); [/asy] $(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2$.

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