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Homeproof of Pythagorean theorem
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proof of Pythagorean theorem
Let $a$ and $b$ be two lengths and let $c$ denote the length of the hypotenuse of a right triangle whose legs have lengths $a$ and $b$.
$\xy,(0,0);(0,40)**@{};(30,0)**@{};(0,0)**@{},(2,15)*{a},(20,2)*{b},(21,16% )*{c}$ 
Behold the following two ways of dissecting a square of length $a+b$:
$\xy,(0,30);(0,0)**@{};(70,0)**@{};(70,70)**@{};(0,70)**@{};(0,30)**@{};(4% 0,0)**@{};(70,40)**@{};(30,70)**@{};(0,30)**@{},(2,15)*{a},(2,50)*{b},(7% 2,20)*{b},(72,55)*{a},(20,2)*{b},(55,2)*{a},(15,72)*{a},(50,72)*{b},(21,16)*% {c},(54,21)*{c},(49,54)*{c},(16,49)*{c},(2,72)*{A},(2,2)*{B},(72,2)*{C},(7% 2,72)*{D},(30,72)*{H},(72,40)*{G},(40,2)*{F},(2,30)*{E}$ 
Figure 1
$\xy,(0,0);(0,70)**@{};(70,70)**@{};(70,0)**@{};(0,0)**@{},(0,40);(70,40)**% @{},(30,0);(30,70)**@{},(0,40);(30,0)**@{},(30,70);(70,40)**@{},(15,72)*{a% },(50,72)*{b},(15,2)*{a},(50,2)*{b},(2,20)*{b},(2,55)*{a},(72,20)*{b},(72,% 55)*{a},(28,55)*{a},(32,20)*{b},(15,42)*{a},(50,38)*{b},(14,19)*{c},(51,56)*{c% },(2,72)*{A^{{\prime}}},(2,2)*{B^{{\prime}}},(72,2)*{C^{{\prime}}},(72,72)% *{D^{{\prime}}},(30,72)*{K},(72,40)*{L},(30,2)*{M},(2,40)*{N},(32,38)*{O}$ 
Figure 2
We now discuss the construction of these diagrams and note some facts about them, starting with the first one. To construct figure 1, we proceed as follows:

Construct a square $ABCD$ whose sides have length $a+b$.

Lay point $E$ on the line segment $AB$ at a distance $a$ from $B$. Likewise, lay the point $F$ on the line segment $BC$ at a distance $a$ from $C$, lay the point $G$ on the line segment $CD$ at a distance $a$ from $D$, and lay the point $H$ on the line segment $DA$ at a distance $a$ from $A$.

Connect the line segments $EF$, $FG$, $GH$, and $HE$.
We now note the following facts about figure 1:

Since the lengths of the four sides of the square $ABCD$ all equal $a+b$ and the line segments $EB$, $FC$, $GD$, and $HA$ were constructed to have length $a$, it follows that the lengths of the line segments $AE$, $BF$, $CG$, $HD$ all equal $b$, as indicated upon the figure.

Since $ABCD$ is a square, the angles $ABC$, $BCD$, $CDA$, and $DAB$ are all right angles, and hence equal each other.

By definition of $c$ as the length of a hypotenuse, it follows that the line segments $EF$, $FG$, $GH$, and $HE$ all have length $c$, as indicated in the figure.

Because the sum of the angles of a triangle equals two right angles and the angle $EAH$ is a right angle, it follows that the sum of the angles $AEH$ and $AHE$ equals a right angle. Since the triangle $AEH$ is congruent to the triangle $BFE$, the angles $AHE$ and $BEF$ are equal. Hence the sum of the angles $AEH$ and $BEF$ equals a right angle. Thus, we may conclude that $HEF$ is a right angle. By similar reasoning, we conclude that $EFG$, $FGH$, and $GHE$ are also right angles.

Since its sides are of equal length and its angles are all right angles, the quadrilateral $EFGH$ is a square.
To construct figure 2, we proceed as follows:

Construct a square $A^{{\prime}}B^{{\prime}}C^{{\prime}}D^{{\prime}}$ whose sides have length $a+b$.

Lay point $N$ on the line segment $A^{{\prime}}B^{{\prime}}$ at a distance $a$ from $A^{{\prime}}$. Likewise, lay the point $M$ on the line segment $B^{{\prime}}C^{{\prime}}$ at a distance $a$ from $B^{{\prime}}$, lay the point $L$ on the line segment $C^{{\prime}}D^{{\prime}}$ at a distance $a$ from $D^{{\prime}}$, and lay the point $K$ on the line segment $D^{{\prime}}A^{{\prime}}$ at a distance $a$ from $A^{{\prime}}$.

Connect the line segments $KM$, $NL$, $MN$, and $KL$.
We now note the following facts about figure 2:

Since the lengths of the four sides of the square $A^{{\prime}}B^{{\prime}}C^{{\prime}}D^{{\prime}}$ all equal $a+b$ and the line segments $A^{{\prime}}N$, $A^{{\prime}}K$, $B^{{\prime}}M$, and $D^{{\prime}}L$ were constructed to have length $a$, it follows that the lengths of the line segments $B^{{\prime}}N$, $C^{{\prime}}L$, $C^{{\prime}}M$, $D^{{\prime}}K$ all equal $b$, as indicated upon the figure.

Since $A^{{\prime}}B^{{\prime}}C^{{\prime}}D^{{\prime}}$ is a square, the angles $A^{{\prime}}B^{{\prime}}C^{{\prime}}$, ${}^{{\prime}}B^{{\prime}}C^{{\prime}}D^{{\prime}}$, $C^{{\prime}}D^{{\prime}}A^{{\prime}}$, and $D^{{\prime}}A^{{\prime}}B^{{\prime}}$ are all right angles, and hence equal each other.

Since $A^{{\prime}}B^{{\prime}}$ and $C^{{\prime}}D^{{\prime}}$, as opposite sides of a square, are parallel and their subsegments $A^{{\prime}}N$ and $D^{{\prime}}L$ have equal length, it follows that $NL$ is parallel to $A^{{\prime}}D^{{\prime}}$ and $B^{{\prime}}C^{{\prime}}$. Likewise, since $A^{{\prime}}D^{{\prime}}$ and $B^{{\prime}}C^{{\prime}}$, as opposite sides of the same square, are parallel and their subsegments $A^{{\prime}}K$ and $B^{{\prime}}M$ have equal length, it follows that $KM$ is parallel to $A^{{\prime}}B^{{\prime}}$ and $C^{{\prime}}D^{{\prime}}$.

Since $NL$ is parallel to $B^{{\prime}}C^{{\prime}}$ and $A^{{\prime}}B^{{\prime}}C^{{\prime}}$ is a right angle, it follows that $A^{{\prime}}NL$ is a right angle; moreover, since $A^{{\prime}}D^{{\prime}}C^{{\prime}}$ is a right angle, $C^{{\prime}}LN$ is also a right angle. Likewise, since $MK$ is parallel to $C^{{\prime}}D^{{\prime}}$ and $A^{{\prime}}D^{{\prime}}C^{{\prime}}$ is a right angle, it follows that $A^{{\prime}}KM$ is a right angle; moreover, since $A^{{\prime}}B^{{\prime}}C^{{\prime}}$ is a right angle, $C^{{\prime}}MK$ is also a right angle. Since $A^{{\prime}}B^{{\prime}}$ is parallel to $KM$ and $A^{{\prime}}NL$ is a right angle, it follows that $KOL$, $LOM$, $MON$, and $NOK$ are all right angles.

Since all the angles of the quadrilateral $A^{{\prime}}KON$ are right angles and two of its adjacent sides, $A^{{\prime}}K$ and $A^{{\prime}}N$, have the same length $a$, this figure is a square, hence the remaining sides, $OK$ and $ON$, also have length $a$. Likewise, Since all the angles of the quadrilateral $C^{{\prime}}LOM$ are right angles and two of its adjacent sides, $C^{{\prime}}L$ and $C^{{\prime}}M$, have the same length $b$, this figure is a square, hence the remaining sides, $OL$ and $OM$, also have length $b$.

By the sideangleside theorem, the traingles $KOL$, $LD^{{\prime}}K$, $MB^{{\prime}}N$, and $NOM$ are congruent to each other and to the triangles $HAE$, $EBF$, $FCG$, and $GDH$.
Hence, we have shown that a square with sides of length $a+b$ may be dissected into either

a square $A^{{\prime}}KON$ with sides of length $a$,

a square $C^{{\prime}}LOM$ with sides of length $b$,

four right triangles, $KOL$, $LD^{{\prime}}K$, $MB^{{\prime}}N$, and $NOM$ with sides of length $a$, $b$, $c$
or

a square $EFGH$ with sides of length $c$,

four right triangles, $HAE$, $EBF$, $FCG$, and $GDH$, with sides of length $a$, $b$, $c$.
Since the area of the whole equals the sum of the areas of the parts and the squares $ABCD$ and $A^{{\prime}}B^{{\prime}}C^{{\prime}}D^{{\prime}}$ are congruent, hence have equal areas, it follows that the sum of the areas of the figures comprising the former dissection equals the sum of the areas of the figures comprising the latter dissection. Since the triangles involved in these dissections all are congruent, hence have equal areas, we may cancel their areas to conclude that the area of $EFGH$ equals the sum of the areas of $A^{{\prime}}KON$ and $C^{{\prime}}LOM$, or $a^{2}+b^{2}=c^{2}$.
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