Download Inscribed Angles Properties

G.C.A.2 STUDENT NOTES WS #6/#7 – geometrycommoncore.com
1
Inscribed Angles Properties
An inscribed angle is an angle that has its vertex on the circle and its
sides contain chords of the circle. The intercepted arc is the arc
that lies in the interior of the inscribed angle and has endpoints on
the angle.
Inscribed ABC and intercepted AC .
B
B
C
A
C
A
Intercepted Arc
Intercepted Arc
Theorem – If an angle is inscribed in a circle, then its measure is half
the measure of its intercepted arc (or half the central angle on that
same intercepted arc).
1
1
mAC  mABC
mADC  mABC
2
2
mADC  2mABC
B
B
1
2
x°
D
D
x°
A
C
A
C
x°
Proof of the Theorem
B
By inserting radius DB we split ABC into two parts, mABC  x  o . Also two
isosceles triangles are formed. The vertex angles of the two isosceles triangle are
180  2x and 180  2o . Thus to determine the mADC we use:
mADC  mADB  mCDB  360
Thus establishing that
mADC  (180  2 x)  (180  2o)  360
mADC = 2mABC.
mADC  2 x  2o
x o
D
C
A
Theorem – If two inscribed angles of a circle intercept the same arc, then those two
angles are congruent.
B
D
Proof of Theorem -- An inscribed angle is half of its intercepted arc.
1
1
mABC  m AC and mADC  mAC and so mADC  mABC .
2
2
A
C
Theorem – If one side of an inscribed triangle is a diameter, then the angle opposite it is
a right triangle.
b)
F
G
Proof of Theorem -- When an angle subtends a diameter, it also subtends an arc of 180.
The inscribed angle is half of its arc which is 90.
EXAMPLES Find the angles & arcs
1a)
o
x
E
c)
1
2
1
1
38°
2
2
114°
m1 = 19 half of central 
m2 = 38 equal to central 
m1 = 57 half of arc
m2 = 114 equal to arc
134°
100°
18°
m1 = 36 double the inscribed 
m2 = 45 (360 – 100 – 36 – 134)/2
G.C.A.2 STUDENT NOTES WS #6/#7 – geometrycommoncore.com
2
Theorem – If a quadrilateral is inscribed in a circle, then opposite angles are
supplementary.
B
A
Proof of Theorem -- The quadrilateral is made up of 4 inscribed angles. We will look
specifically at DAB and DCB.
D
C
mDAB =
1
mDCB
2
mDCB =
B
1
mDB
2
1
1
mDCB  mDB
2
2
1
mDAB  mDCB  mDCB  mDB
2
1
mDAB  mDCB   360 
2
mDAB  mDCB  180
mDAB  mDCB 
B

A
A
D
D

C
C
We would use the same technique to prove that mADC  mABC  180 .
EXAMPLES -- Find the requested angle and arc values.
a)
b)
c)
118°
64°
1
1
258°
58°
1
m1 = 116
102°
2
180 – 64
m2 = 122 180 - 58
2
2
m1 = 90 inscribed on diameter
m2 = 62
180 - 118
m1 = 51 half of central 
m2 = 129 half of central 