Free Math Printable Worksheets: Circles Worksheets and Activites for Math Teachers

### Angles formed by intersecting Chords

**Theorem: **The measure of the angle formed by two chords that intersect **ins****ide** the circle is $$ \frac{1}{2}$$ the sum of the chords' intercepted arcs.

**Note:** This theorem applies to the angles and arcs of any chords that intersect within the circle. It is __NOT__ necessary for these chords to intersect at the center of the circle for this theorem to apply.

In the diagram on the left, the measure of x is half the sum of the intercepted arcs ($$ \overparen{ABC} $$ and $$ \overparen{DFG} $$)

**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

$$ \angle A= \frac{1}{2} \cdot ( \text{sum of intercepted arcs }) \\ \angle A= \frac{1}{2} \cdot ( \overparen{HIJ} + \overparen{ KLM } ) \\ \angle A= \frac{1}{2} \cdot ( 38^ {\circ} + 68^ {\circ}) \\ \angle A= \frac{1}{2} \cdot ( 108 ^{\circ} ) \\ \angle A= 54 ^{\circ} $$

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.