Unizor - Geometry2D - Area - Pythagorean Theorem
Channel: Zor Shekhtman on UNIZOR Education
Date: 16/10/2012

Proof 1 (by Pythagoras)
Let our right triangle have catheti a, b and hypotenuse c.
Construct a square ABCD with a side equal to a+b and choose four points, one on each side of this square:
E on AB at distance a from A,
F on BC at distance a from B,
G on CD at distance a from C,
H on DA at distance a from D.
Connect these points sequentially into a figure EFGH. Based on congruency of four right triangles in the corners of an original square (by two catheti), it's easy to proove that EFGH is a square with a side equal to a hypotenuse c. Therefore, its area is c^2 and the area of an original square with a side a+b equals to four areas of a right triangle with catheti a, b and hypotenuse c plus c^2.
Now let's shift our corner triangles.
Move ΔFCG parallel to itself towards ΔEAH such that hypotenus FG will lie over hypotenuse EH. Vertex F will go over E, vertex G will go over H and vertex C will move to a position we mark as X.
Move ΔEBF parallel to itself along line BC until its vertex F coincides with point C. Vertex E then will go over point X.
Move ΔGDH parallel to itself along line CD until its vertex G coincides with point C. Vertex H then will go over point X.
Now we still have the same four right triangles inside an original square, but the rest of this square is a combination of two squares with sides, correspondingly, a and b and, therefore, areas of a^2 and b^2.
Since, shifting triangles around we did not deform anything, the areas that complement four of these triangle to an original square does not change. Therefore, c^2 = a^2 + b^2.
End of Pythagoras' proof.

Proof 2 (algebraic method)
Use the same arrangement as in the previous proof. The area of an original square is (a+b)^2. The area of each right triangle in the corner of this square is a·b/2. Therefore, we can write an equation:
(a+b)^2 = 4·a·b/2 + c^2.
Simplifying left and right parts, we get:
a^2 + 2·a·b + b^2 = 2·a·b + c^2.
Therefore, a^2 + b^2 = c^2.
End of proof.

Proof 3 (by Euclid)
Consider a right triangle ΔABC with point A being a vertex at the right angle and BC - a hypotenuse (label vertices counterclockwise for definiteveness).
Construct squares on each side of this triangle (all outside of it):
AHKC on cathetus AC,
ABFG on cathetus AB,
BCED on hypotenus BC.
Draw an altitude from vertex A intersecting hypotenuse BC at point M and continue it until it intersect the other side of square BCED at point L of its side ED.
We will show that the area of square AHKC is equal to the area of rectangle MCEL and the area of square ABFG is equal to the area of rectangle BMLD. This will be the proof that sum of squares of catheti in a right triangle equals to a square of a hypotenuse.
Now, triangle ΔAKC has an area equal to half of an area of square AHKC and also has an area equal to that of triangle ΔBCK (because both ΔAKC and ΔBCK share the base CK and have the same altitudes - the distance between parallel lines BH and CK).
Similarly, triangle ΔCEL has an area equal to half of an area of rectangle MCEL and also has an area equal to that of triangle ΔACE (because both ΔCEL and ΔACE share the base CE and have the same altitudes - the distance between parallel lines AL and CE).
So, we have shown that half the area of square AHKC equals to an area of triangle ΔBCK and half the area of rectangle MCEL equals to an area of triangle ΔACE. But triangles ΔBCK and ΔACE are congruent by side-angle-side. Therefore, areas of square AHKC and rectangle MCEL are equal.
Exactly the same way we can prove that areas of square ABFG and rectangle BMLD are equal.
That concludes Euclid's proof.

Proof 4 (using similarity and algebra)
Consider a right triangle with catheti a, b and hypotenuse c. Let an altitude from a vertex of a right angle to a hypotenus be h and segments it divides a hypotenuse be x and y (so, x + y = c). An original triangle and two triangles it is divided into by an altitude from a vertex of a right angle are, obviously, similar because their accute angles are congruent.
Therefore, we can write the following equations of proportionality:
b/h = a/x,
a/h = b/y,
a/h = c/b.
Getting x and y from the first two equations and h from the third one, we obtain:
x = a·h/b,
y = b·h/a,
h = a·b/c.
Substitute h from the last equation into the previous two:
x = a^2/c,
y = b^2/c.
Add the last two equations together and replace x+y with c:
c = (a^2+b^2)/c or
c^2 = a^2+b^2.
End of proof.