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9.04 Inscribed angles and polygons

Lesson

Concept summary

Angles formed by chords of a circle are known as inscribed angles:

Inscribed angle in a circle

An angle which is formed in the interior of a circle when two chords share an endpoint.

Two chords are on a circle which both have an endpoint at the same point on the circle. The angle between the chords is an inscribed angle in the circle.

Multiple inscribed angles which share segments can form an inscribed polygon:

Inscribed polygon in a circle

A polygon which has all of its vertices on a circle.

Quadrilateral C E D B inscribed in circle A.

When a polygon is inscribed in a circle, we can equivalently say that the circle is circumscribed on the polygon.

A circle can also be inscribed into a polygon:

Inscribed circle in a polygon

The largest possible circle that can be drawn in the interior of a polygon. If it is a regular polygon, then each side of the polygon is tangent to the circle.

A circle inscribed in a regular hexagon.

Again, we can refer to this equivalently by saying that the polygon is circumscribed on the circle.

The following theorems relate to angles and polygons inscribed in circles:

Inscribed angle theorem

If an angle is inscribed in a circle, then its measure is half the measure of its intercepted arc.

An angle with a measure of theta degrees inscribed in a circle. The angle intercepts the arc A B with a measure of 2 theta degrees.
Congruent inscribed angle theorem

If two inscribed angles of a circle intercept the same arc, then the angles are congruent.

A circle with points A, C, D, and B placed clockwise on a circle. Chords A C, A D, C B, and B D. Chords A D and C B intersect at a point. Angles A C B and A D B are congruent.
Inscribed semicircle theorem

An angle inscribed in a semicircle is a right angle.

Circle A with diameter B D, chords B C and C D. Angle B C D is a right angle.
Inscribed right triangle theorem

If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle.

Circle A with diameter B C. A right triangle is inscribed in circle A with B C as the hypotenuse of the triangle.
Opposite inscribed angle theorem

The opposite angles of an inscribed quadrilateral are supplementary.

Quadrilateral B C D E inscribed in circle A. Angles B and E are marked.
Quadrilateral B C E D inscribed in circle A.

The opposite inscribed angle theorem says that two opposite angles in a quadrilateral inscribed in a circle add up to 180 \degree.

In this case, that results in the following equations:

m\angle BCE + m\angle EDB = 180

m\angle DBC + m\angle CED = 180

Worked examples

Example 1

Given m\angle CEB = 4x + 11 and m\angle CDB = 12x - 5. Find m\angle CDB

 Circle A with points C, E, D, and B placed clockwise on the circle. Chords C E, C D, C B, E B, and D B are drawn.

Approach

By the congruent inscribed angle theorem, we know m\angle CEB = m\angle CDB. We want to write an equation relating the two angles and then solve for x.

Solution

\displaystyle 4x + 11\displaystyle =\displaystyle 12x - 5
\displaystyle 4x + 16\displaystyle =\displaystyle 12xAdd 5 to both sides
\displaystyle 16\displaystyle =\displaystyle 8xSubtract 4x from both sides
\displaystyle 2\displaystyle =\displaystyle xDivide both sides by 8
\displaystyle x\displaystyle =\displaystyle 2Symmetric property of equality

We have established x = 2, so we can substitute that into the equation for \angle CDB.

Substituting x = 2 into 12x - 5 we get 12\left(2\right) - 5 = 19. Therefore m\angle CDB = 19 \degree

Reflection

Note that since m\angle CDB = m\angle CEB, we could have used the expression for m\angle CEB to calculate the size of the angle instead:

\displaystyle m\angle CEB\displaystyle =\displaystyle 4x + 11Given
\displaystyle =\displaystyle 4\left(2\right) + 11Substitute x = 2
\displaystyle =\displaystyle 19Simplify the expression

which is the same result.

Example 2

Solve for x.

Angle A T B with a measure of negative 10 plus x degrees inscribed in a circle. Angle A T B intercepts an arc A B which has a measure of 4 plus x degrees.

Approach

We can use the inscribed angle theorem to write an equation and then solve for x.

Solution

\displaystyle 2(m\angle{ATB}) \displaystyle =\displaystyle m \overset{\large\frown}{AB}
\displaystyle 2(-10+x)\displaystyle =\displaystyle 4+xSubstitution
\displaystyle -20+2x\displaystyle =\displaystyle 4+xDistribute the 2
\displaystyle 2x\displaystyle =\displaystyle 24+xAdd 20 to both sides
\displaystyle x\displaystyle =\displaystyle 24Subtract x from both sides

x=24

Reflection

The value of x is 24, which means the inscribed angle m \angle ATB = 14 \degree, and the intercepted arc m\overset{\large\frown}{AB} = 28 \degree.

Outcomes

MA.912.GR.5.1

Construct a copy of a segment or an angle.

MA.912.GR.5.2

Construct the bisector of a segment or an angle, including the perpendicular bisector of a line segment.

MA.912.GR.5.4

Construct a regular polygon inscribed in a circle. Regular polygons are limited to triangles, quadrilaterals and hexagons.

MA.912.GR.6.2

Solve mathematical and real-world problems involving the measures of arcs and related angles.

MA.912.GR.6.3

Solve mathematical problems involving triangles and quadrilaterals inscribed in a circle.

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