Download All angles inscribed in a circle that subtend the same chord are equal

Euclid’s Proposition 111.21
(One of its results)
This theorem is a result of Euclid’s Proposition 111.21: In a circle the angles in
the same segment equal one another. It is a simple result with a fairly straightforward
proof. It is easy to prove this proposition using Euclid’s Proposition 111.20: In a circle
the angle at the center is double the angle at the circumference when the angles have the
same circumference as base. This is a simple way to prove 111.21 since if we know the
angle at the center is double the angle at the circumference, with the same circumference
as their bases; we basically know a center angle is double the same inscribed angle that
subtends the same chord. Therefore if we have two inscribed angles that subtend the
same chord, they are both equal to one half of the same center angle and therefore must
be equal to one another.
Hence the theorem is proven! But that is the easy way out. Instead, first let us
prove proposition 111.21 to show that proposition 111.20 is also true.
To prove the Central Angle Theorem (111.21, as it is often called) we must
consider three cases. Begin by assuming the angle, <ABC, is an inscribed angle of circle
L, with center O.
Case 1: One of the sides of <ABC goes through O. (See Figure 1) Since OA,
OB, and OC are radii of circle L and therefore have equal length, by the Isosceles
Triangle Theorem, m<OBA = m<OAB and m<OAC = m<OCA. Then, since we know
all triangles have angles that sum to 180 degrees, *2m<OAB + 2m<OAC = 180 degrees.
This is the same as adding m<OBA + m<OAB + m<OAC + m<OCA = 180 degrees. We
also know **m<AOC = 180º - 2m<OAC. Substituting * into ** at 180, we have
EMAT 6690
Cynthia Thomas
Page 1 of 5
m<AOC = 2m<OAB + 2m<OAC – 2m<OAC
 m<AOC = 2m<OAB.
Since the inscribed angle <ABC = m<OBA and the central angle <AOC = 2m<OAB, we
have proven a central angle is double an inscribed angle where they share the same base
(AC in this particular case).
A
B
O
C
Figure 1
Case 2: O lies in the interior of <ABC (or ∆ABC). (See Figure 2) We construct a
point D, which lies on L and BO. We know m<ABC = m<ABD + m<CBD and m<ABC
= m<AOD + m<COD. Since OB, OD, OA, and OC are radii of L, we know they have
equal measure. Therefore m<OAB = m<OBA, m<OAD = m<ODP, m<OCB = m<OBC,
and m<OCD = m<ODC, by isosceles triangles. Since the angles of a triangle sum to 180
degrees, we have m∆BAD = 2m<OAB + 2m<OAD = 180º and m∆BCD = 2m<OCB +
2m<OCD = 180º. We also know m<AOD = 180º - 2m<OAD and <COD = 180º 2m<OCD. By substitution,
m<AOD = 2m<OAB + 2m<OAD – 2m<OAD
 m<AOD = 2m<OAB, since m<OAB = m<OBA,
EMAT 6690
Cynthia Thomas
Page 2 of 5
 m<AOD = 2m<OBA.
And,
m<COD = 2m<OCB + 2m<OCD – 2m<OCD
 m<COD = 2m<OCB, since m<OCB = m<OBC,
 m<COD = 2m<OBC.
We also have that
<ABC = <ABD + <CBD = <OBA + <OBC and <AOC = <AOD + <COD.
Which means <AOC = 2m<OBA + 2m<OBC from our results above. Therefore <AOC
= 2(m<OBA + m<OBC). Since <ABC and <AOC both subtend AC, and <AOC is a
central angle, we have proven the theorem for Case 2.
A
B
D
O
C
Figure 2
Case 3: O is neither on <ABC or in the interior of the triangle ABC (See Figure
3). Construct a point T such that T lies on BO, this creates a line segment BOT or BT.
We know m<ABC = m<ABT – m<CBT and m<AOC = m<AOT – m<COT. To simplify
the notation, let a = m<OBA, b = m<OTA, c = m<OBC, and d = m <OTC. Since OA,
OB, OC and OT are all radii of the circle and thereby have equal measure, by the
EMAT 6690
Cynthia Thomas
Page 3 of 5
Isosceles Triangle Theorem, m<OAB = m<OBA = a, m<OAT = m<OTA = b, m<OCB =
m<OBC =c, and m<OCT = m<OTC = d. By the Angle Sum Theorem, the m∆BAT = 2a
+ 2b = 180º and m∆BCT = 2c + 2d = 180º. By the same theorem, m<AOT = 180º - 2b
and m<COT = 180º - 2d. By substitution, m<AOT = 2a – 2b – 2b and m<COT = 2c + 2d
– 2d. This gives:
m<AOT = 2a and m<COT = 2c.
Since m<ABC = m<ABT – m<CBT = m<OBA – m<OBC = a – c and m<AOC =
m<AOT – m<COT = 2a – 2c = 2(a-c), we have two angles (<ABC and <AOC) that
subtend the same chord, AC, and one of the angles (<AOC) is a central angle. And the
measure of the central angle, <AOC is equal to two times the measure of the other angle,
<ABC. Therefore the theorem is proven.
A
C
B
O
T
Figure 3
By the above proof of the Central Angle Theorem it is very easy to see how two
inscribed angles that subtend the same chord have the same measure. Figure 4 shows this
by the construction of a central angle sharing the same chord that the two inscribed
angles share.
EMAT 6690
Cynthia Thomas
Page 4 of 5
Reference
www.duke.edu/~kfr/Scans/CATPaper.doc
EMAT 6690
Cynthia Thomas
Page 5 of 5