﻿ Formula for Angles of intersecting chords theorem. Example and practice problems. First off you need to ...

# Angles of intersecting chords theorem

Formula for angles and intercepted arcs of intersecting chords

### 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

$$x = \frac{1}{2} \cdot ( \text{sum of intercepted arcs }) \\ x = \frac{1}{2} \cdot ( \overparen{TG } + \overparen{ ER } ) \\ x = \frac{1}{2} \cdot ( 75^ {\circ} + 65^ {\circ}) \\ x = \frac{1}{2} \cdot ( 140 ^{\circ} ) \\ x= 70 ^{\circ}$$

Note: $$\overparen {JK}$$ is not an intercepted arc , so it cannot be used for this problem.

By our theorem, we know that

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

Note: $$\overparen { NO }$$ is not an intercepted arc , so it cannot be used for this problem.

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 angles

The 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.

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