PROPOSITION 8.
In a given square to inscribe a circle.
Let
ABCD be the given square; thus it is required to inscribe a circle in the given square
ABCD.
Let the straight lines
AD,
AB be bisected at the points
E,
F respectively [
I. 10], through
E let
EH be drawn parallel to either
AB or
CD, and through
F let
FK be drawn parallel to either
AD or
BC; [
I. 31] therefore each of the figures
AK,
KB,
AH,
HD,
AG,
GC,
BG,
GD is a parallelogram, and their opposite sides are evidently equal. [
I. 34]
Now, since
AD is equal to
AB, and
AE is half of
AD, and
AF half of
AB,
therefore AE is equal to AF, so that the opposite sides are also equal;
therefore FG is equal to GE.
Similarly we can prove that each of the straight lines
GH,
GK is equal to each of the straight lines
FG,
GE;
therefore the four straight lines GE, GF, GH, GK are equal to one another.
Therefore the circle described with centre
G and distance one of the straight lines
GE,
GF,
GH,
GK will pass also through the remaining points.
And it will touch the straight lines
AB,
BC,
CD,
DA, because the angles at
E,
F,
H,
K are right.
For, if the circle cuts
AB,
BC,
CD,
DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle : which was proved absurd; [
III. 16] therefore the circle described with centre
G and distance one of the straight lines
GE,
GF,
GH,
GK will not cut the straight lines
AB,
BC,
CD,
DA.
Therefore it will touch them, and will have been inscribed in the square
ABCD.
Therefore in the given square a circle has been inscribed. Q. E. F.
PROPOSITION 9.
About a given square to circumscribe a circle.
Let
ABCD be the given square; thus it is required to circumscribe a circle about the square
ABCD.
For let
AC,
BD be joined, and let them cut one another at
E.
Then, since
DA is equal to
AB, and
AC is common, therefore the two sides
DA,
AC are equal to the two sides
BA,
AC; and the base
DC is equal to the base
BC;
therefore the angle DAC is equal to the angle BAC. [I. 8]
Therefore the angle
DAB is bisected by
AC.
Similarly we can prove that each of the angles
ABC,
BCD,
CDA is bisected by the straight lines
AC,
DB.
Now, since the angle
DAB is equal to the angle
ABC, and the angle
EAB is half the angle
DAB, and the angle
EBA half the angle
ABC, therefore the angle
EAB is also equal to the angle
EBA; so that the side
EA is also equal to
EB. [
I. 6]
Similarly we can prove that each of the straight lines
EA,
EB is equal to each of the straight lines
EC,
ED.
Therefore the four straight lines
EA,
EB,
EC,
ED are equal to one another.
Therefore the circle described with centre
E and distance one of the straight lines
EA,
EB,
EC,
ED will pass also through the remaining points; and it will have been circumscribed about the square
ABCD.
Let it be circumscribed, as
ABCD.
Therefore about the given square a circle has been circumscribed. Q. E. F.
