PROPOSITION 22.
If there be three plane angles of which two,
taken together in any manner,
are greater than the remaining one,
and they are contained by equal straight lines,
it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines.
Let there be three plane angles
ABC,
DEF,
GHK, of which two, taken together in any manner, are greater than the remaining one, namely
the angles ABC, DEF greater than the angle GHK, the angles DEF, GHK greater than the angle ABC, and, further, the angles
GHK,
ABC greater than the angle
DEF; let the straight lines
AB,
BC,
DE,
EF,
GH,
HK be equal, and let
AC,
DF,
GK be joined; I say that it is possible to construct a triangle out of straight lines equal to
AC,
DF,
GK, that is, that any two of the straight lines
AC,
DF,
GK are greater than the remaining one.
Now, if the angles
ABC,
DEF,
GHK are equal to one another, it is manifest that,
AC,
DF,
GK being equal also, it is possible to construct a triangle out of straight lines equal to
AC,
DF,
GK.
But, if not, let them be unequal, and on the straight line
HK, and at the point
H on it, let the angle
KHL be constructed equal to the angle
ABC; let
HL be made equal to one of the straight lines
AB,
BC,
DE,
EF,
GH,
HK, and let
KL,
GL be joined.
Now, since the two sides
AB,
BC are equal to the two sides
KH,
HL, and the angle at
B is equal to the angle
KHL, therefore the base
AC is equal to the base
KL. [
I. 4]
And, since the angles
ABC,
GHK are greater than the angle
DEF, while the angle
ABC is equal to the angle
KHL, therefore the angle
GHL is greater than the angle
DEF.
And, since the two sides
GH,
HL are equal to the two sides
DE,
EF, and the angle
GHL is greater than the angle
DEF, therefore the base
GL is greater than the base
DF. [
I. 24]
But
GK,
KL are greater than
GL.
Therefore
GK,
KL are much greater than
DF.
But
KL is equal to
AC; therefore
AC,
GK are greater than the remaining straight line
DF.
Similarly we can prove that
AC,
DF are greater than
GK, and further
DF,
GK are greater than
AC.
Therefore it is possible to construct a triangle out of straight lines equal to
AC,
DF,
GK. Q. E. D.
