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

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. Produce

C A

to

F

in the direction of

A

(Postulate 2), and lay off

A F

on

C A

produced so that

E F = E B

(I.3, taking

E B

as the standard length). On

A F

describe the square

F G H A

(I.46); produce

G H

to meet

C D

at

K

. Then by II.6 applied to

C F

bisected at

A

with extension

A F

, the rectangle on

C F

,

F A

together with the square on

E A

equals the square on

E F

. But

E F = E B

, so this rectangle plus square on

E A

equals the square on

E B

, which by I.47 (in

△ A B E

, right-angled at

A

) equals the square on

E A

plus the square on

A B

. Subtracting the square on

E A

from both sides (Common Notion 3): \[ CF \cdot FA \;=\; AB^2. \] The rectangle

C K

on

C F

,

F A

(=

C F \cdot F G

since

F G = F A

) equals the square on

A B

. Subtracting the common rectangle on

F A

,

A H

from both, the square

F G H A

on

F A

equals the rectangle

H K

on

H D

and

D K = A B - A H

. Setting

A H = A F

on

A B

(point

H

on

A B

with

A H = A F

) gives the desired section:

A B \cdot H B = A H^{2}

.

Figure

Proposition II.11. Square $ABDC$ on $AB$; midpoint $E$ of $AC$;
$F$ on the extension of $AC$ with $EF = EB$. Then $AH = AF$ cuts $AB$
in the desired ratio: $AB \cdot HB = AH^2$. This is the golden section. — Proposition II.11. Square $A B D C$ on $A B$ ; midpoint $E$ of $A C$ ; $F$ on the extension of $A C$ with $E F = E B$ . Then $A H = A F$ cuts $A B$ in the desired ratio: $A B \cdot H B = A H^{2}$ . This is the golden section.

lines 74–74 in main.tex