PROPOSITION 23.
To construct a solid angle out of three plane angles two of which,
taken together in any manner,
are greater than the remaining one: thus the three angles must be less than four right angles.
Let the angles
ABC,
DEF,
GHK be the three given plane angles, and let two of these, taken together in any manner, be greater than the remaining one, while, further, the three are less than four right angles; thus it is required to construct a solid angle out of angles equal to the angles
ABC,
DEF,
GHK.
Let
AB,
BC,
DE,
EF,
GH,
HK be cut off equal to one another, and let
AC,
DF,
GK be joined; it is therefore possible to construct a triangle out of straight lines equal to
AC,
DF,
GK. [
XI. 22]
Let
LMN be so constructed that
AC is equal to
LM,
DF to
MN, and further
GK to
NL, let the circle
LMN be described about the triangle
LMN, let its centre be taken, and let it be
O; let
LO,
MO,
NO be joined; I say that
AB is greater than
LO.
For, if not,
AB is either equal to
LO, or less.
First, let it be equal.
Then, since
AB is equal to
LO, while
AB is equal to
BC, and
OL to
OM, the two sides
AB,
BC are equal to the two sides
LO,
OM respectively; and, by hypothesis, the base
AC is equal to the base
LM; therefore the angle
ABC is equal to the angle
LOM. [
I. 8]
For the same reason the angle
DEF is also equal to the angle
MON, and further the angle
GHK to the angle
NOL; therefore the three angles
ABC,
DEF,
GHK are equal to the three angles
LOM,
MON,
NOL.
But the three angles
LOM,
MON,
NOL are equal to four right angles; therefore the angles
ABC,
DEF,
GHK are equal to four right angles.
But they are also, by hypothesis, less than four right angles: which is absurd.
Therefore
AB is not equal to
LO.
I say next that neither is
AB less than
LO.
For, if possible, let it be so, and let
OP be made equal to
AB, and
OQ equal to
BC, and let
PQ be joined.
Then, since
AB is equal to
BC,
OP is also equal to
OQ, so that the remainder
LP is equal to
QM.
Therefore
LM is parallel to
PQ, [
VI. 2] and
LMO is equiangular with
PQO; [
I. 29] therefore, as
OL is to
LM, so is
OP to
PQ; [
VI. 4] and alternately, as
LO is to
OP, so is
LM to
PQ. [
V. 16]
But
LO is greater than
OP; therefore
LM is also greater than
PQ.
But
LM was made equal to
AC; therefore
AC is also greater than
PQ.
Since, then, the two sides
AB,
BC are equal to the two sides
PO,
OQ, and the base
AC is greater than the base
PQ, therefore the angle
ABC is greater than the angle
POQ. [
I. 25]
Similarly we can prove that the angle
DEF is also greater than the angle
MON, and the angle
GHK greater than the angle
NOL.
Therefore the three angles
ABC,
DEF,
GHK are greater than the three angles
LOM,
MON,
NOL.
But, by hypothesis, the angles
ABC,
DEF,
GHK are less than four right angles; therefore the angles
LOM,
MON,
NOL are much less than four right angles.
But they are also equal to four right angles: which is absurd.
Therefore
AB is not less than
LO.
And it was proved that neither is it equal; therefore
AB is greater than
LO.
Let then
OR be set up from the point
O at right angles to the plane of the circle
LMN, [
XI. 12] and let the square on
OR be equal to that area by which the square on
AB is greater than the square on
LO; [Lemma] let
RL,
RM,
RN be joined.
Then, since
RO is at right angles to the plane of the circle
LMN, therefore
RO is also at right angles to each of the straight lines
LO,
MO,
NO.
And, since
LO is equal to
OM, while
OR is common and at right angles, therefore the base
RL is equal to the base
RM. [
I. 4]
For the same reason
RN is also equal to each of the straight lines
RL,
RM; therefore the three straight lines
RL,
RM,
RN are equal to one another.
Next, since by hypothesis the square on
OR is equal to that area by which the square on
AB is greater than the square on
LO, therefore the square on
AB is equal to the squares on
LO,
OR.
But the square on
LR is equal to the squares on
LO,
OR, for the angle
LOR is right; [
I. 47] therefore the square on
AB is equal to the square on
RL; therefore
AB is equal to
RL.
But each of the straight lines
BC,
DE,
EF,
GH,
HK is equal to
AB, while each of the straight lines
RM,
RN is equal to
RL; therefore each of the straight lines
AB,
BC,
DE,
EF,
GH,
HK is equal to each of the straight lines
RL,
RM,
RN.
And, since the two sides
LR,
RM are equal to the two sides
AB,
BC, and the base
LM is by hypothesis equal to the base
AC, therefore the angle
LRM is equal to the angle
ABC. [
I. 8]
For the same reason the angle
MRN is also equal to the angle
DEF, and the angle
LRN to the angle
GHK.
Therefore, out of the three plane angles
LRM,
MRN,
LRN, which are equal to the three given angles
ABC,
DEF,
GHK, the solid angle at
R has been constructed, which is contained by the angles
LRM,
MRN,
LRN. Q. E. F.
LEMMA.
But how it is possible to take the square on
OR equal to that area by which the square on
AB is greater than the square on
LO, we can show as follows.
Let the straight lines
AB,
LO be set out, and let
AB be the greater; let the semicircle
ABC be described on
AB, and into the semicircle
ABC let
AC be fitted equal to the straight line
LO, not being greater than the diameter
AB; [
IV. 1] let
CB be joined
Since then the angle
ACB is an angle in the semicircle
ACB, therefore the angle
ACB is right. [
III. 31]
Therefore the square on
AB is equal to the squares on
AC,
CB. [
I. 47]
Hence the square on
AB is greater than the square on
AC by the square on
CB.
But
AC is equal to
LO.
Therefore the square on
AB is greater than the square on
LO by the square on
CB.
If then we cut off
OR equal to
BC, the square on
AB will be greater than the square on
LO by the square on
OR. Q. E. F.
