Formula for Angles of intersecting chords theorem. Example and practice problems with step by step solutions.

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.

Diagram 1

In diagram 1, the x is half the sum of the measure of the intercepted arcs ($$ \overparen{ABC} $$ and $$ \overparen{DFG} $$)

Note: This theorem applies to the angles and arcs of 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

Problem 1

Chords $$ \overline{JW} $$ and $$ \overline{LY} $$ intersect as shown below. $$\text{m } \overparen{\red{JKL}} $$ is $$ 75^{\circ}$$ $$\text{m } \overparen{\red{WXY}} $$ is $$ 65^{\circ}$$ and What is the value of $$a$$?

example of angle formed by intersecting chords

By our theorem, we know that

$$ a = \frac{1}{2} \cdot (\text{sum of intercepted arcs }) \\ a = \frac{1}{2} \cdot (\text{m } \overparen{\red{JKL}} + \text{m } \overparen{\red{WXY}} ) \\ a = \frac{1}{2} \cdot (75^ {\circ} + 65^ {\circ}) \\ a = \frac{1}{2} \cdot (140 ^{\circ}) \\ a= 70 ^{\circ} $$

Problem 2

If $$ \overparen{\red{HIJ}}= 38 ^{\circ} $$ , $$ \overparen{JK} = 44 ^{\circ} $$ and $$ \overparen{KLM}= 68 ^{\circ} $$, then what is the measure of $$ \angle $$ A?

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

By our theorem, we know that.

$$ \angle A= \frac{1}{2} \cdot (\text{sum of intercepted arcs }) \\ \angle A= \frac{1}{2} \cdot (\overparen{\red{HIJ}} + \overparen{ \red{KLM } }) \\ \angle A= \frac{1}{2} \cdot (38^ {\circ} + 68^ {\circ}) \\ \angle A= \frac{1}{2} \cdot (106 ^{\circ}) \\ \angle A= 53 ^{\circ} $$

Problem 3

If $$ \overparen{MNL}= 60 ^{\circ}$$, $$ \overparen{NO}= 110 ^{\circ}$$and $$ \overparen{OPQ}= 20 ^{\circ} $$, then what is the measure of $$ \angle Z $$?

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

By our theorem, we know that.

$$ \angle Z= \frac{1}{2} \cdot (\text{sum of intercepted arcs }) \\ \angle Z= \frac{1}{2} \cdot (\color{red}{ \overparen{ NML }}+ \color{red}{\overparen{ OPQ } }) \\ \angle Z= \frac{1}{2} \cdot (60 ^ {\circ} + 20^ {\circ}) \\ \angle Z = \frac{1}{2} \cdot (80 ^{\circ}) \\ \angle Z= 40 ^{\circ} $$

Problem 4

Use the theorem for intersecting chords to find the value of sum of intercepted arcs (assume all arcs to be minor arcs).

picture of chords intersecting in circle

By our theorem, we know that.

$$ 110^{\circ} = \frac{1}{2} \cdot (\text{sum of intercepted arcs }) \\ 110^{\circ} = \frac{1}{2} \cdot (\overparen{TE } + \overparen{ GR }) \\ 2 \cdot 110^{\circ} =2 \cdot \frac{1}{2} \cdot (\overparen{TE } + \overparen{ GR }) \\ 220 ^{\circ} =\overparen{TE } + \overparen{ GR } $$

Problem 5

Find the measure of AEB and CED.

$$ \angle AEB = \frac{1}{2} (\overparen{ AB} + \overparen{ CD}) \\ \angle AEB = \frac{1}{2}(30 ^{\circ} + 25 ^{\circ}) \\ \angle AEB = \frac{1}{2} (55 ^{\circ}) \\ \angle AEB = 27.5 ^{\circ} $$

$$ m \angle AEB = m \angle CED$$ CED since they are vertical angles

Problem 6

What is wrong with this problem, based on the picture below and the measurements?

$ m \angle AEC = 70 ^{\circ} \\ \overparen{AGF}= 170 ^{\circ } \\ \overparen{CD}= 40 ^{\circ } $

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.