
Related Pages:
 Circles forumula, graph, equations
 Equation of A Circle
 Circumference
 Area
 Chord
 Tangent
 Arc of A Circle
 Intersecting Chords
 Inscribed Angle
 Secant of circle
 2 Tangents from 1 point
 Central Angle
 Angles, Arc, Secants, tangents
 Tangents, Secants and Side Lengths
 Tangent and a Chord
 images
Angles formed by intersecting Chords
Theorem: :
The measure of the angle formed by 2 chords
that intersect inside the circle is $$ \frac{1}{2}$$ the sum of the chords' intercepted arcs.
In the diagram on the left, the measure of x is half the sum of the intercepted arc ($$ \overparen{ABC} $$ and $$ \overparen{DFG} $$)
Note: This theorem applies to the angles and arcs of any chords that intersect anywhere within the circle. It is not necessary for these chords to intersect at the center of the circle for this theorem to apply.
Practice Problem
By our theorem, we know that
$$ x = \frac{1}{2} \cdot ( \text{sum of intercepted arcs }) \\ x = \frac{1}{2} \cdot ( \overparen{TG } + \overparen{ ER } ) \\ x = \frac{1}{2} \cdot ( 75^ {\circ} + 65^ {\circ}) \\ x = \frac{1}{2} \cdot ( 140 ^{\circ} ) \\ x= 70 ^{\circ} $$
By our theorem, we know that
$$ \angle A= \frac{1}{2} \cdot ( \text{sum of intercepted arcs }) \\ \angle A= \frac{1}{2} \cdot ( \overparen{HIJ} + \overparen{ KLM } ) \\ \angle A= \frac{1}{2} \cdot ( 38^ {\circ} + 68^ {\circ}) \\ \angle A= \frac{1}{2} \cdot ( 106 ^{\circ} ) \\ \angle A= 53 ^{\circ} $$
By our theorem, we know that
$$ \angle Z= \frac{1}{2} \cdot ( \text{sum of intercepted arcs }) \\ \angle Z= \frac{1}{2} \cdot (\color{red}{ \overparen{ NML }}+ \color{red}{\overparen{ OPQ } }) \\ \angle Z= \frac{1}{2} \cdot ( 60 ^ {\circ} + 20^ {\circ}) \\ \angle Z = \frac{1}{2} \cdot ( 80 ^{\circ} ) \\ \angle Z= 40 ^{\circ} $$
By our theorem, we know that
$$ 110^{\circ} = \frac{1}{2} \cdot ( \text{sum of intercepted arcs }) \\ 110^{\circ} = \frac{1}{2} \cdot ( \overparen{TE } + \overparen{ GR } ) \\ 2 \cdot 110^{\circ} =2 \cdot \frac{1}{2} \cdot ( \overparen{TE } + \overparen{ GR } ) \\ 220 ^{\circ} =\overparen{TE } + \overparen{ GR } $$
$$ \angle AEB = \frac{1}{2} ( \overparen{ AB} + \overparen{ CD}) \\ \angle AEB = \frac{1}{2}(30 ^{\circ} + 25 ^{\circ}) \\ \angle AEB = \frac{1}{2} (55 ^{\circ}) \\ \angle AEB = 27.5 ^{\circ} $$
$$ m \angle AEB = m \angle CED$$ CED since they are vertical anglesThe problem with these measurements is that if angle AEC = 70°, then we know that $$\overparen{ ABC }$$ + $$\overparen{ DF }$$ should equal 140°.
So, there are two other arcs that make up this circle. Namely, $$ \overparen{ AGF }$$ and $$ \overparen{ CD }$$. These two other arcs should equal 360°  140° = 220°
So far everything is fine. However, the measurements of $$ \overparen{ CD }$$ and $$ \overparen{ AGF }$$do not add up to 220°
170 + 40 ≠220
Therefore, the measurements provided in this problem violate the theorem that angles formed by intersecting arcs equals the sum of the intercepted arcs.

Related Pages:
 Circles forumula, graph, equations
 Equation of A Circle
 Circumference
 Area
 Chord
 Tangent
 Arc of A Circle
 Intersecting Chords
 Inscribed Angle
 Secant of circle
 2 Tangents from 1 point
 Central Angle
 Angles, Arc, Secants, tangents
 Tangents, Secants and Side Lengths
 Tangent and a Chord
 images