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the bisector of an angle

We introduced the bisector of an angle in (8) as follows
If P,Q,R are distinct points, then there is a unique line through Q making equal angles with
rays QP and QR (through R if R is on the ray QP) this is called the bisector of <PQR.
We gave this clumsy description in order to include cases where R lies on the line PQ.
Here, we shall examine the cases in some more detail.
(1) R lies on the line QP, but not on the ray QP - so Q is between P and R.
Then the rays QP, QR make angle π.
A line through Q which makes equal angles with both must make angle π/2 with each
This must be the perpendicular to PQ at Q.
(2) R is on the ray QP.
Then the rays QP and QR coincide, so make zero angle.
There now appear to be two candidates - the line QP and the perpendicular to QP at Q.
We added the condition "through R" to select the former.

(3) R is not on the line QP.
We need to know how the line can be drawn.
The sketch on the right shows a typical configuration.
By one of Hilbert's axioms, we can choose S on the ray QR with n(Q,S) = n(Q,P).
By (7) the segment PS has a midpoint T, so we have n(P,T) = n(T,S).
Now ΔQTP and ΔQTS have corresponding sides equal, so we can use (SSS) to show
that the ray QT is equally inclined to QP and QR as <PQT = <TQR.
Thus, QT is the required line.
NOTE the congruence also shows that <QTP = <QTS, so PS is perpendicular to QT.
We shall need the following result:

Lemma
Suppose that the line L is the bisector of <PQR. Then r_L(P) lies on the ray QR.

proof
In case (1), L is the line through Q perpendicular to QP.
Then r_L(P) lies on QP, since this is perpendicular to L, and lies
on the opposite side of L to P. Thus it lies on the same side as R,
and so on the ray QR.
In case (2), L is the line QP, so r_L(P) = P, and so is on QR.
In case (3), L is the line QT above. Now, we chose S as a point on the ray QR.
We showed also that QT is perpendicular to PS. Since also n(P,T) = n(T,S), we
see that r_L(P) = S, so is on the ray QR.

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We introduced the bisector of an angle in (8) as follows If P,Q,R are distinct points, then there is a unique line through Q making equal angles with rays QP and QR (through R if R is on the ray QP) this is called the bisector of <PQR. We gave this clumsy description in order to include cases where R lies on the line PQ. Here, we shall examine the cases in some more detail. (1) R lies on the line QP, but not on the ray QP - so Q is between P and R. Then the rays QP, QR make angle π. A line through Q which makes equal angles with both must make angle π/2 with each This must be the perpendicular to PQ at Q. (2) R is on the ray QP. Then the rays QP and QR coincide, so make zero angle. There now appear to be two candidates - the line QP and the perpendicular to QP at Q. We added the condition "through R" to select the former.
(3) R is not on the line QP. We need to know how the line can be drawn. The sketch on the right shows a typical configuration. By one of Hilbert's axioms, we can choose S on the ray QR with n(Q,S) = n(Q,P). By (7) the segment PS has a midpoint T, so we have n(P,T) = n(T,S). Now ΔQTP and ΔQTS have corresponding sides equal, so we can use (SSS) to show that the ray QT is equally inclined to QP and QR as <PQT = <TQR. Thus, QT is the required line. NOTE the congruence also shows that <QTP = <QTS, so PS is perpendicular to QT. We shall need the following result:
Lemma Suppose that the line L is the bisector of <PQR. Then r_L(P) lies on the ray QR. proof In case (1), L is the line through Q perpendicular to QP. Then r_L(P) lies on QP, since this is perpendicular to L, and lies on the opposite side of L to P. Thus it lies on the same side as R, and so on the ray QR. In case (2), L is the line QP, so r_L(P) = P, and so is on QR. In case (3), L is the line QT above. Now, we chose S as a point on the ray QR. We showed also that QT is perpendicular to PS. Since also n(P,T) = n(T,S), we see that r_L(P) = S, so is on the ray QR.