12.04 Inscribed angles and polygons
An inscribed angle is an angle with a vertex is on the circumference of a circle and legs that contain the chords of a circle. The arc that lies in the interior of the inscribed angle is called the intercepted arc.
Exploration
In the applet below $\angle DFE$∠DFE is the inscribed angle and arc $DE$DE is its intercepted arc. Move points $D$D, $E$E, and $F$F around the circle and try to answer the following questions.
- What relationship exists between the measure of the inscribed angle and its intercepted arc?
- What has to happen for $\angle F$∠F to be a right angle?
From the applet, we can see that the inscribed angle is always half the measure of its intercepted arc. Since we can prove this is true, it's known as the inscribed angle theorem.
If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc.
In the diagram below, $a=\frac{1}{2}b$a=12b.
The measure of an angle is half the measure of the intercepted arc, so $a=\frac{1}{2}b$a=12b.
In the applet above, we saw that if we move the point on the circumference of the circle, while keeping the arc the same, the measure of the inscribed angle remains the same. This is a corollary to the inscribed angle theorem.
If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent.
In the diagram below, $\angle ADB$∠ADB and $\angle BCA$∠BCA intercept the same arc. Therefore, $\angle ADB\cong\angle BCA$∠ADB≅∠BCA.
Angles intercepting the same arc are congruent.
Inscribed polygons
We say that a polygon is inscribed in a circle if all of its vertices are on the circumference of the circle. The polygon is sometimes referred to as an inscribed polygon.
Triangles and quadrilaterals that are inscribed in circles have special properties. First, let's think about the relationship between a semicircle and an inscribed right angle.
Exploration
Suppose that an inscribed angle has a measure of $90^\circ$90°. Then it follows from the inscribed angle theorem that its intercepted arc is $2\times90^\circ=180^\circ$2×90°=180°. An arc measuring $180^\circ$180° is a semicircle. Therefore, an inscribed right angle intercepts a semicircle and the two chords form a diameter of the circle. We can prove the converse in a similar way.
An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.
For the diagram below, $m\angle ACB=90^\circ$m∠ACB=90° if and only if $\overline{AB}$AB is a diameter of the circle.
Since $\overline{AB}$AB is a diameter of circle $O$O, $\angle C$∠C is a right angle.
The following theorem also follows from the inscribed angle theorem. As an exercise, see if you can prove its truth in a two-column or paragraph proof.
If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.
For example, in quadrilateral $ABCD$ABCD below, we know that $m\angle A+m\angle C=180^\circ$m∠A+m∠C=180° and $m\angle B+m\angle D=180^\circ$m∠B+m∠D=180°.
In quadrilateral $ABCD$ABCD, opposite angles are supplementary.
Practice questions
Question 1
Solve for $x$x.
Question 2
Find the value of $x$x.
Question 3
In the diagram, $O$O is the center of the circle.
A quadrilateral is inscribed in a circle, with its top side being the circle's diameter. A diagonal is drawn from the top-left vertex to the bottom-right vertex of the quadrilateral, forming two triangles, one on top of the other. In the top triangle, the sides are the diameter of the circle, the diagonal of the quadrilateral and the right side of the quadrilateral. The interior angles of the top triangle are labeled. The top-left angle is the angle between the diameter and the diagonal, and is labeled as $27^\circ$27°, indicating its measure. The top-right angle is the angle between the diameter and the right side, and is labeled $\left(q\right)$(q). The bottom-right angle is the angle between the diagonal and the right side, and is labeled as $\left(p\right)$(p). The $\left(p\right)$(p) bottom-right angle is subtended by the diameter. The bottom-left angle of the quadrilateral is labeled as $\left(r\right)$(r). The $\left(r\right)$(r) bottom-left angle is opposite the $\left(q\right)$(q) top-right angle of the triangle.
Solve for $p$p.
Solve for $q$q.
Solve for $r$r.