Pythagorean theorem: a^2+b^2=c^2

tl;dr; If you aren't eager to learn proof of the pythagorean theorem, you can stop reading.

I hope, almost everybody knows or has heard about pythagorean theorem. I am not aware in which year of school it is taught, but it is definitely in school syllabus. In this post, I am going to write proof of it.

Let's recall pythagorean theorem. If we have right triangle(triangle with a 90-degree angle) ABC , then this equation is true:

a^2 + b^2 = c^2

or in other denotations:

|BC|^2 + |AC|^2 = |AB|^2

1. Proof by area

We want to proof that a^2+b^2=c^2.

Area of the outside rectangle is equal to (a+b)^2. But, there is a other way to calculate outside area, we can sum up area of the inside rectangle and four right triangles, which are located on the corners.

(a+b)^2 = c^2 + (a*b)/2 * 4

if we open the bracket on the left side,

a^2+2*a*b+b^2=c^2+2*a*b

subtract 2*a*b from both sides,

a^2+b^2=c^2

End of the proof

2. Proof by triangle

Try to come up, why angle at Point C is divided into b and a ? (btw, it is called similar triangles).

sin(a) = opposite/hypotenuse = |BC| / |AB| = |BD| / |BC|

cos(a) = adjacent/hypotenuse = |AC| / |AB| = |AD| / |AC|

in other words,

|BC|^2 = |AB| * |BD| - derived by cross multiplying the sin(a) proportion above.

|AC|^2= |AB| * |AD| - derived by cross multiplying the cos(a) proportion above.

if we sum up two equations,

|BC|^2+|AC|^2=|AB|*(|AD|+|BD|)

From the picture we can verify that:

|AD|+|BD|=|AB|

Thus,

|BC|^2+|AC|^2=|AB|^2

End of the proof.