Solutions and Comments for Problems in Book I
Exercise 1. Given a segment AB and its midpoint M, show that the distance
CM is one half the difference between CA and CB if C is a point on the segment.
If C is on line AB, but not between A and B, then CM is one half the sum of CA
Solution. If point C lies between M and B (Figure t1a), we have AM = AC −
MC and MB = MC + CB. Since AM = MB, we have AC − MC = MC + CB,
so that MC =
(AC − BC). A similar argument holds if C lies between A and M.
If point C lies on the extension of segment AB past point B (Figure t1b), then
AM = MB leads to AC − MC = MC − BC, so that MC =
(AC + BC), and
similarly if point C lies on the extension of segment AB past point A.
Notes. It is a useful exercise for students, using dynamic geometry software,
to compute the measure of AC − BC and AC + BC as point C slides along line
AB. The result—that the sum is constant inside the segment while the difference
is constant (in absolute value) outside the segment—is demonstrated dramatically.
Exercise 2. Given an angle AOB and its bisector OM, show that angle COM
is one half the difference of COA and COB if ray OC is inside angle AOB; it is
the supplement of half the difference if ray OC is inside A OB , which is vertical
to AOB; and it is one half the sum of COA and COB if OM is inside one of the
other angles AOA and BOB formed by these lines.
Solution. If ray OC1 lies inside MOB (Figure t2), then AOM = MOB =
C1OA − C1OM = C1OB + C1OM, so that C1OM =
(C1OA − C1OB), with a
similar result if ray OC1 lies inside AOM.
If ray OC2 lies inside A OB , then its extension OC1 lies inside AOB, and
MOC2 = 180◦ − MOC1, and the required result follows.
If ray OC3 (not shown in diagram) lies inside BOA , then AOC3 − MOC3 =
MOC3 − BOC3, so that MOC3 =
(AOC3 + BOC3), and similarly if ray OC3 lies
inside AOB .