constructing & proving a
square
Proposition 46 -
from Book 1 of The Elements
[Bonus material: proving the regular pentagon!]
Brought to you by: François, Frank & Zack
Euclid: Book 1, Proposition 46
“To Construct A Square With A Given Side”
Constructing the Square
Step 1
Let AB be the given straight line (Postulate 1).
A
B
Step 2
Draw AC at right angles to given line AB from point A (Proposition 11).
C
A
B
Step 3
Make a point D on line AC so that AD=AB (Proposition 3).
C
D
A
B
Step 4
Draw DE from point D parallel to AB (Proposition 31).
C
E
D
A
B
Step 5
Draw BE from point B Parallel to AD (Proposition 31) .
C
D
E
A
B
Step 6 and Step 7
Since ADEB is a parallelogram, AB=DE and AD=BE (Proposition 34) .
Since in “Step 2” we said lines AD=AB, then lines AD=DE=AB=BE make
parallelogram (Common Notion 1).
C
E
D
A
B
ADEB an equilateral
Step 8
Since line AD falls upon parallels AB and DE, BAD+ ADE=2 right angles (Proposition 29).
C
E
D
A
B
Step 9
Since BAD is a right angle as stated in “Step 1”,
ADE must also be another right angle.
C
E
D
A
B
Step 10
In “parallelogrammic” areas, the opposite sides and angles are =, so each of the opposite angles of
ABE and BED are also right. As a result, ADEB is right angled (Proposition 34).
C
E
D
A
B
Step 11
Since we proved ADEB is equilateral and right angled, it is therefore a square, and it is described
along line AB (Definition 22).
C
E
D
A
B
The Pentagon! (Book IV, Proposition 11)
Given: Circle
ABCDE
F
A
**Where triangle FGH
is isosceles and where
angles G and H are
each double that size
of angle F. (Book 4,
Prop 10)
E
B
D
C
G
H
(1)
Inscribe triangle FGH into the circle, where triangle ACD is equiangular with triangle FGH (Book IV
Prop. 2)
-
F
A
E
B
D
C
G
H
Bisect the angles ACD and ADC using straight lines CE and DB.
A
Angles ACD and CDA are
double the size of angle
CAD.
E
B
-But they have been
bisected, therefore these
five angles, DAC, ACE,
ECD, CDB and BDA are
all equal to each other.
C
D
Circumferences
Equal angles
stand on equal
circumferences
, so all five
circumferences
are equal (AB,
BC, CD, DE,
EA). (Book III,
Prop. 26)
AB, BC, CD, DE, EA are equal
A
E
B
C
D
Equilateral as Lines AB, BC, CD, DE, and EA are equal.
Straight lines that
cut off equal
circumferences are
equal, so lines AB,
BC, CD, DE, and EA
are equal. (Book III,
Prop. 29)
Therefore
pentagon ABCDE
is equilateral.
A
E
B
C
D
Equiangular
Circumference AB = DE
Add BCD to each so:
AB + BCD = DE + BCD (CN 2)
A
Therefore ABCD = EDCB.
Angle AED is on circumference
ABCD, and angle BAE is on
circumference EDCB (Book III,
Prop 27) therefore angle BAE =
angle AED.
Same applies for the rest
of the angles so all angles
are equal and it is
Equiangular.
E
B
C
D
Equilateral Pentagon
+
Equiangular Pentagon
=
Regular Pentagon
THE END