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

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II.14 Proposition II.14

To construct a square equal to a given rectilineal figure.

Proof

Let

A

be the given rectilineal figure. By Proposition I.45, construct a parallelogram

B C D E

equal in area to

A

, with the parallelogram's angles right (apply I.46 on one of its sides if necessary so it is a rectangle). If

B E = E D

then

B C D E

is already a square and the construction is complete. Suppose

B E > E D

; the case

B E < E D

is symmetric. Produce

B E

to

F

, laying off

E F = E D

on the produced line (I.3). Bisect

B F

at

G

(I.10). With centre

G

and radius

GB = GF

describe the semicircle

B H F

above

B F

. Produce

D E

to meet the semicircle at

H

. Join

G H

; then

G H

is a radius and equals

GF

. Apply II.5 to

B F

bisected at

G

and cut at

E

:

B E \cdot E F + E G^{2} = G F^{2} = G H^{2}

. By I.47 in the right triangle

GE H

(right angle at

E

because

E H ⊥ B F

):

G H^{2} = E G^{2} + E H^{2}

. Subtract

E G^{2}

from both expressions (Common Notion 3):

B E \cdot E F = E H^{2}

. But

E F = E D

, so

B E \cdot E D = E H^{2}

; the square on

E H

equals the rectangle

B C D E

, which equals the original figure

A

.

Figure

$Proposition II.14. Reduce the given rectilineal figure to rectangle $BCDE$ (I.45). Extend $BE$ to $F$ with $EF = ED$; bisect $BF$ at $G$; describe the semicircle on $BF$. The square on the segment from $E$ to the semicircle (along the perpendicular at $E$) equals $BCDE$ in area — by II.5 + I.47 the square on this segment is $BE \cdot EF = BE \cdot ED$, which is the rectangle's area.$

Proposition II.14. Reduce the given rectilineal figure to rectangle

B C D E

(I.45). Extend

B E

to

F

with

E F = E D

; bisect

B F

lines 74–74 in main.tex

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