Source — Euclid's Elements, encoded as an rrxiv paper

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paper/figures/fig-i-1.pdfimage · 2409 bytesRaw

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draw

C GF

parallel to either

A D

or

B E

(I.31), meeting

B D

at

G

and

D E

at

F

. Through

G

draw

H K

parallel to either

A B

or

D E

(I.31), meeting

A D

at

H

and

B E

at

K

. Since

C F

is parallel to

A D

and

B D

falls on them, the exterior angle

∠ B GC

equals the interior and opposite

∠ B D A

(I.29). But

∠ B D A = ∠ D B A

since

B A = A D

(I.5 applied to the isoceles right triangle inside the square). Hence

∠ B GC = ∠ GB C

, so

B C = C G

(I.6), and therefore

C B K G

is equilateral. Since it has a right angle at

B

, it is a square on

C B

(Definition I.22). By the same reasoning

H D F G

is the square on

H D = A C

. The complements

A G H D

and

GF B K

in the square

A D E B

are equal rectangles by I.43; each is contained by

A C

and

C B

(since

A H = A C

,

H G = C B

, etc.), so each is the rectangle on

A C

,

C B

. The four pieces sum to the whole (Common Notion 2): \[ AB^2 \;=\; AC^2 + CB^2 + 2\cdot(AC\cdot CB), \] which is

(a + b)^{2} = a^{2} + 2 ab + b^{2}

in geometric form.