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II.4 Proposition II.4

If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.

Proof

Let the straight line

A B

be cut at random at

C

. Describe on

A B

the square

A D E B

(I.46), and draw the diagonal

B D

. Through

C

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.

Figure

$Proposition II.4. The square $ADEB$ on $AB = a+b$ is decomposed by the parallels $CF \parallel AD$ and $HK \parallel AB$ into the square $HDFG$ on $AC = a$, the square $CBKG$ on $CB = b$, and two equal rectangles $AGHD$ and $GFBK$ (by I.43), each equal to $a \cdot b$.$

Proposition II.4. The square

A D E B

on

A B = a + b

is decomposed by the parallels

C F ∥ A D

and

H K ∥ A B

lines 74–74 in main.tex

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II.4 Proposition II.4

If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.

Proof

Let the straight line

A B

be cut at random at

C

. Describe on

A B

the square

A D E B

(I.46), and draw the diagonal

B D

. Through

C

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.

Figure

lines 74–74 in main.tex

into the square

H D F G

on

A C = a

, the square

C B K G

on

C B = b

, and two equal rectangles

A G H D

and

GF B K

(by I.43), each equal to

a \cdot b

.