Absolute geometry
Circles (properties)
Properties are taken from Kay, p 105. Fill in the missing items – note that if a fill in number repeats within
a problem (e.g. you see [Fill in (2)] in two places in the same property), it means that that exact statement is
being repeated.
Property 1:
The center of a circle is the midpoint of any diameter.
Proof:
Statement: Consider a circle centered at O with diameter PQ .
Reason: Given.
Statement: Because P and Q are points on the circle, segments OP and OQ are both radii
of the circle. Therefore:
[Fill in (1)]
(What can you say about their lengths?)
Reason: Definition of radius.
Statement: Since PQ is a diameter, O  PQ (definition), and
[Fill in (2)]
(What betweenness relationship holds?)
Statement: Since [Fill in (1) from above] and [Fill in (2) from above], O is the [Fill in (3)]
of segment PQ .
Reason: Definition of a [Fill in (3) from above].
Property 2:
The perpendicular bisector of any chord of a circle passes through the center.
Proof:
Statement: Let a circle centered at O with radius r, and with
chord PQ given. Let RS be the perpendicular bisector of PQ .
(Be careful with appearances here – you don't know that
RS passes through O, although it certainly appears to. That's
the thing you're trying to prove.)
Reason: Given.
Now, there's a theorem on p 144 you want to look at. By this
theorem, the perpendicular bisector of PQ is the set of all
points which are an equal distance from the endpoints P and Q. So, if any one point can
be shown to be an equal distance from P and Q, it must be on the perpendicular bisector.
Statement: OP = r = OQ
Reason: Since P and Q lie on the circle and O is the center, segments OP and OQ are
both [Fill in (1)] of that circle, and thus have equal length.
Statement: Therefore, since O is an equal distance from P and Q, it must lie on the
perpendicular bisector (i.e., the perpendicular bisector passes through the center).
Reason: [Fill in (2) – name that theorem!]
Property 3:
A line passing through the center and perpendicular to a chord bisects the chord.
Proof:
Statement: Let a circle centered at O with radius r, and with
chord PQ given. Let RS be the perpendicular to PQ , and
passing through point O.
Reason: Given.
Statement: Since O is the center, and P and Q are points on the
circle, segments OP and OQ are both [Fill in (1)] of the circle
and therefore [Fill in (2)]
Reason: (The reason is simply by definition of that thing.)
Statement: Since RS  PQ , PTO and
PTO and QTO are right triangles).
QTO are both [Fill in (3)], (and therefore
Reason: [Fill in (4)]
Statement: OT  OT .
Reason: [Fill in (5)]
Statement: PTO 
QTO
Reason: One of those right triangle congruence criteria on pp. 176-177. Which one? [Fill
in (6)]
Statement: [Fill in (7)]
Reason: CPCF
Statement: And so T is the midpoint of PQ .
Reason: [Fill in (8)]
Statement: Which gives RS bisects PQ .
Reason: [Fill in (9)]
Property 4:
Two congruent central angles subtend congruent chords, and conversely.
Note: The proof of this appears in the text as Example 1 on page 195, but it's worth
taking apart in a bit more detail. First off, notice the "and conversely" – this is two
separate statements, or a version of the "if and only if."
What it is saying is that
(1) If you have two congruent central angles, then the chords subtended by them
are congruent.
And
(2) If you have two congruent chords, then the central angles which subtend them
are congruent.
These two statements need to be proven separately.
Proof of (1):
Statement: Let a circle centered at O with radius r, points A, B, C, D
on the circle. Suppose DOC  AOB (note DOC subtends
chord DC , AOB subtends chord AB ).
Reason: Given.
Statement: OD, OC , OB, and OA are all [Fill in (1)] of the circle, and
therefore [Fill in (2)]
Reason: By definition.
Statement: So we can say about triangles DOC and AOB that [Fill in (3)]
Reason: [Fill in (4)]
Statement: And therefore [Fill in (5)]
Reason: [Fill in (6)]
Proof of (2):
Statement: Let a circle centered at O with radius r, points A, B, C,
D on the circle. Suppose DC  AB (note DOC subtends
chord DC , AOB subtends chord AB ).
Reason: Given.
Statement: OD, OC , OB, and OA are all [Fill in (1) (as above)] of
the circle, and therefore [Fill in (2) (as above)]
Reason: By definition.
Statement: So we can say about triangles DOC and AOB that [Fill in (3)(as above)]
Reason: [Fill in (7)]
Statement: And therefore [Fill in (8)]
Reason: [Fill in (9)]
Property 5:
Two chords equidistant from the center of a circle have equal lengths, and
conversely.
Proof: This is proven as Example 2 on p 195, and is also posted as a live example.
Key things to watch for are that (1) this is another "if and only if" proof – it has to be
done in both directions, and (2) the distance from one geometric object to another is
defined to be the shortest (perpendicular) distance. In particular, the distance from a point
to a line is defined on page 179. I don't think we've encountered that before.
PS – you don't need to do anything for this one. :)