42 Distance; degree-measure of an angle Ch. 3
A
B
C
H 1
k
Figure 3.7. Laying off an angle.
A
B
C
B 1
C 1
Figure 3.8. Opposite angles at a vertex.
Degree-measure has the properties:-
(i)The null-angle∠BAB has degree-measure0.
(ii)For any non-null wedge-angle∠BAC, we have 0 <|∠BAC|◦<180.
(iii)If∠B 1 AC 1 is the angle vertically opposite to∠BAC, then
|∠B 1 AC 1 |◦=|∠BAC|◦,
so that vertically opposite angles have equal degree-measures.
Proof.
(i) LetCbe a point not onAB.ThenbyA 5 (iii) withD=B,
|∠BAB|◦+|∠BAC|◦=|∠BAC|◦.
It follows that|∠BAB|◦=0.
(ii) Given any non-null wedge-angle∠BAC,letH 1 be the closed half-plane with
edgeABin whichClies. If we had|∠BAC|◦=0, then we would have|∠BAC|◦=
|∠BAB|◦andsobyA 5 (iv) we would have[A,C=[A,B. This would imply that∠BAC
is null, contrary to assumption. Then by A 5 (i)|∠BAC|◦>0.
Now choose the pointE=Aso thatA∈[B,E].ThenbyA 5 (iii), as we have sup-
plementary angles,
|∠BAC|◦+|∠CAE|◦= 180.
But[A,E =[A,C as∠BACis a wedge-angle, so∠CAEis not a null-angle. By the
last paragraph we then have|∠CAE|◦>0 and it follows that|∠BAC|◦<180.
(iii) As|BAB 1 ,|CAC 1 are straight we have
|∠BAC|◦+|∠CAB 1 |◦= 180 ,
|∠CAB 1 |◦+|∠B 1 AC 1 |◦= 180 ,
there being two pairs of supplementary angles. It follows that
|∠BAC|◦+|∠CAB 1 |◦=|∠CAB 1 |◦+|∠B 1 AC 1 |◦,
from which we conclude by subtraction that|∠BAC|◦=|∠B 1 AC 1 |◦.