PROPOSITION 13.
In a given pentagon,
which is equilateral and equiangular,
to inscribe a circle.
Let
ABCDE be the given equilateral and equiangular pentagon; thus it is required to inscribe a circle in the pentagon
ABCDE.
For let the angles
BCD,
CDE be bisected by the straight lines
CF,
DF respectively; and from the point
F, at which the straight lines
CF,
DF meet one another, let the straight lines
FB,
FA,
FE be joined.
Then, since
BC is equal to
CD, and
CF common, the two sides
BC,
CF are equal to the two sides
DC,
CF; and the angle
BCF is equal to the angle
DCF;
therefore the base BF is equal to the base DF, and the triangle
BCF is equal to the triangle
DCF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [
I. 4]
Therefore the angle
CBF is equal to the angle
CDF.
And, since the angle
CDE is double of the angle
CDF, and the angle
CDE is equal to the angle
ABC, while the angle
CDF is equal to the angle
CBF; therefore the angle
CBA is also double of the angle
CBF;
therefore the angle ABF is equal to the angle FBC; therefore the angle
ABC has been bisected by the straight line
BF.
Similarly it can be proved that the angles
BAE,
AED have also been bisected by the straight lines
FA,
FE respectively.
Now let
FG,
FH,
FK,
FL,
FM be drawn from the point
F perpendicular to the straight lines
AB,
BC,
CD,
DE,
EA.
Then, since the angle
HCF is equal to the angle
KCF, and the right angle
FHC is also equal to the angle
FKC,
FHC,
FKC are two triangles having two angles equal to two angles and one side equal to one side, namely
FC which is common to them and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [
I. 26] therefore the perpendicular
FH is equal to the perpendicular
FK.
Similarly it can be proved that each of the straight lines
FL,
FM,
FG is also equal to each of the straight lines
FH,
FK; therefore the five straight lines
FG,
FH,
FK,
FL,
FM are equal to one another.
Therefore the circle described with centre
F and distance one of the straight lines
FG,
FH,
FK,
FL,
FM will pass also through the remaining points; and it will touch the straight lines
AB,
BC,
CD,
DE,
EA, because the angles at the points
G,
H,
K,
L,
M are right.
For, if it does not touch them. but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle: which was proved absurd. [
III. 16]
Therefore the circle described with centre
F and distance one of the straight lines
FG,
FH,
FK,
FL,
FM will not cut the straight lines
AB,
BC,
CD,
DE,
EA;
therefore it will touch them.
Let it be described, as
GHKLM.
Therefore in the given pentagon, which is equilateral and equiangular, a circle has been inscribed. Q. E. F.
