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Suggested problems - solutions
Circular arc measure
Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, 2001. In particular, see section 3.8, pp 203-207.
The problems are all from section 3.8.
Problems: 2, 3, 11
#2:
Find the measures of x and y of the two arcs in the figure, then find mABC and check
the additive property in this case.
The first thing I’d do is put some more points in the figure so I can properly talk about the arcs
and angles. Assume betweenness and ordering of points as shown on the diagram.
Now, x and y are easy to justify:
Since APB is a minor arc, x = mAPB = m∠AOB = 143.
Since BQC is a minor arc, y = mBQC = m∠BOQ = 127.
The reason the next bit is a nuisance is because the problem asks you to “check the additive property”. If we were just going to use the additive property, we’d say that since ABC = APB ∪BQC,
with only point B in common, mABC = mAPB + mBQC = x + y = 143 + 127 = 270, and be
done with it.
The checking bit is presumably against the angles, though, and it’s a bit of a nuisance; ABC is
a major arc, formed by the exterior of ∠AOC. Its measure is 360 − m∠AOC, and we don’t
have m∠AOC.
We also, tempting as it is, don’t have a theorem that states that non-intersecting central angles
of a circle sum to 360. You could state and prove a general one (it follows from linear pairs,
which is what I’m about to do), though you might find coming up with the theorem statement
to be a bit tricky (how many central angles are you carving out of your circle, and can you write
a general formula?).
Try this:
−→
−→
Extend ray OB in the opposite direction to form OT . Then angles ∠AOB and ∠AOT form a
linear pair and are supplementary by the linear pair axiom, so m∠AOT = 37.
Also, angles ∠BOC and ∠COT form a linear pair and are supplementary by the linear pair
axiom, so m∠COT = 53.
By angle addition, m∠AOC = m∠AOT + m∠COT = 37 + 53 = 90.
And since ABC is a major arc formed by the exterior of ∠AOC, its measure is 360 − m∠AOC =
360 − 90 = 270. This checks with the result of additivity of arcs.
#3:
Prove two chords are congruent iff they subtend arcs of equal measure.
Note that’s an “if and only if.” Break into two separate statements:
(1) If two chords subtend arcs of equal measure, then the chords are congruent.
(2) If two chords are congruent, then they subtend arcs of equal measure.
It’s also worth a note that by definition, we only talk about chords subtending minor arcs, so we
don’t have to worry about the cases of major arcs or semicircles.
Proof of (1):
Given a circle centered at O with point A, B, C, D, P , Q as shown, suppose that chords AB
and CD subtend minor arcs of equal measure: mAPB = mCQD.
The measures of the central angles equal the measures of their associated arcs, so m∠AOB =
m∠COD and ∠AOB ∼
= ∠COD.
And by circle property four, congruent central angles subtend congruent chords, so AB ∼
= CD.
Proof of (2):
Given a circle centered at O with point A, B, C, D, P , Q as shown, suppose that chords
AB ∼
= CD.
By circle property four, congruent chords are subtended by congruent central angles, so ∠AOB ∼
=
∠COD.
And therefore, the minor arcs subtended by these angles have equal measure: mAPB = mCQD.
#11:
A circle passes through the vertices of ABCD, and AB = CD. Prove that m∠A =
m∠D.
Although the hint for this one says use arcs, and I moved it to this problem set accordingly...I
really don’t see why you’d want to. You need to draw in the radii to talk about the central
angles, and once you’ve done that bit, you’ve got congruent triangles simply from the radii...
Since OA, OB, OC, and OD are all radii of the circle, they are all congruent (I could’ve use the
same tick marks for all those lengths, in fact. So AOB ∼
= COD by the SSS Theorem. And
while we’re at it, AOD is isosceles with base AD.
So, by CPCF, and the Isosceles Triangle Theorem, we have
∠BAO ∼
= ∠CDO, and ∠OAD ∼
= ∠ODA.
Then (assuming betweenness as it appears, which you generally want to do with the circle
problems - the point ordering is as shown, because it would take forever to spell it all out), by
angle addition
m∠BAD = m∠BAO + m∠OAD
m∠CDA = m∠CDO + m∠ODA
and substitution gives m∠BAD = m∠CDA, or m∠A = m∠D.