A circle is inscribed in an isosceles triangle with legs of

In Chapter 2, we saw that the sum of the measures of the interior angles of a polygon with n sides is given by . In turn, the measure I of each interior angle of a regular polygon of n sides is given by . The sum of the measures of the exterior angles of any polygon is always 360°. Thus, the measure E of each exterior I = n - 2180 n S = n - 2180 A B D C Figure 7.26 Solution Figure 7.25b: The center of a circumscribed circle must lie at the same distance from each vertex of the hexagon. Center X is the point of concurrency of the perpendicular bisectors of two consecutive sides of the hexagon. In Figure 7.25b, we construct the perpendicular bisectors of and to locate point X. Figure 7.25c: Where XM is the distance from X to vertex M, we use radius to construct circumscribed . } X XM NP MN 쮿 For a rectangle, which is not a regular polygon, we can only circumscribe the cir- cle see Figure 7.26. Why? For a rhombus also not a regular polygon, we can only inscribe the circle see Figure 7.27. Why? As we shall see, we can construct both inscribed and circumscribed circles for reg- ular polygons because they are both equilateral and equiangular. A few of the regular polygons are shown in Figure 7.28. H K L J Figure 7.27 Equilateral Triangle Figure 7.28 Square Regular Pentagon Regular Octagon Exs. 1–6 EXAMPLE 3 a Find the measure of each interior angle of a regular polygon with 15 sides. b Find the number of sides of a regular polygon if each interior angle measures 144°. Solution a Because all of the n angles have equal measures, the formula for the measure of each interior angle, becomes which simplifies to 156°. I = 15 - 2180 15 I = n - 2180 n angle of a regular polygon of n sides is . E = 360 n b Because , we can determine the number of sides by solving the equation Then NOTE: In Example 3a, we could have found the measure of each exterior angle and then used the fact that the interior angle is its supplement. With leads to . It follows that or 156. In Example 3b, I = 180° - 24° E = 24° E = 360 ° n n = 15, n = 10 36n = 360 180n - 360 = 144n n - 2180 = 144n n - 2180 n = 144 I = 144° 쮿 a A D F E B C b A D F E B C 1 2

3 4

O c A D F E B C 1 2

3 4

5 O d O A D F E B C e A D F E B C O f O A D F E B C S R P M Q N Figure 7.29 Regular polygons allow us to inscribe and to circumscribe a circle. The proof of the following theorem will establish the following relationships:

1. The centers of the inscribed and circumscribed circles of a regular polygon

are the same.

2. The angle bisectors of two consecutive angles or the perpendicular bisectors

of two consecutive sides can be used to locate the common center of the inscribed circle and the circumscribed circle.

3. The inscribed circle’s radius is any line segment from the center drawn

perpendicular to a side of the regular polygon; the radius of the circumscribed circle joins the center to any vertex of the regular polygon. Exs. 7, 8 A circle can be circumscribed about or inscribed in any regular polygon. THEOREM 7.3.1 GIVEN: Regular polygon ABCDEF [See Figure 7.29a.] PROVE: A circle O can be circumscribed about ABCDEF and a circle with center O can be inscribed in ABCDEF. the fact that leads to . In turn, becomes which leads to n = 10. 36° = 360° n , E = 360 ° n E = 36° I = 144° PROOF: Let point O be the point at which the angle bisectors for and meet. [See Figure 7.29b on page 340.] Then and . Because by the definition of a regular polygon, it follows that In turn, , so . Then sides opposite of a are also . From the facts that , , and , it follows that by SAS. [See Figure 7.29c.] In turn, by CPCTC, so because these lie opposite and . Because and , it follows that m ∠4 = 1 2 m ∠BCD ∠5 ⬵ ∠4 OD OC ∠4 ⬵ ∠5 OC ⬵ OD 䉭OCB ⬵ 䉭OCD BC ⬵ CD OC ⬵ OC ∠3 ⬵ ∠4 ⬵ 䉭 ∠s ⬵ OB ⬵ OC ∠2 ⬵ ∠3 m ∠2 = m∠3 1 2 m ∠ABC = 1 2 m ∠BCD ∠ABC ⬵ ∠BCD ∠3 ⬵ ∠4 ∠1 ⬵ ∠2 ∠BCD ∠ABC . But because these are angles of a ∠BCD ⬵ ∠CDE m ∠5 = 1 2 m ∠BCD regular polygon. Thus, , and bisects . ∠CDE OD m ∠5 = 1 2 m ∠CDE By continuing this procedure, we can show that bisects , bisects , and bisects . The resulting triangles, , , , , , and , are congruent by ASA. [See Figure 7.29d.] By CPCTC, . With O as center and as radius, circle O can be circumscribed about ABCDEF , as shown in Figure 7.29e. Because corresponding altitudes of are also congruent, we see that , where these are the altitudes to the bases of the triangles. Again with O as center, but now with a radius equal in length to OM, we complete the inscribed circle in ABCDEF. [See Figure 7.29f.] In the proof of Theorem 7.3.1, a regular hexagon was drawn. The method of proof would not change, regardless of the number of sides of the polygon chosen. In the proof, point O was the common center of the circumscribed and inscribed circles for ABCDEF. Because any regular polygon can be inscribed in a circle, any regular polygon is cyclic. OM ⬵ ON ⬵ OP ⬵ OQ ⬵ OR ⬵ OS 䉭s ⬵ OA OA ⬵ OB ⬵ OC ⬵ OD ⬵ OE ⬵ OF 䉭FOA 䉭EOF 䉭DOE 䉭COD 䉭BOC 䉭AOB ∠FAB OA ∠EFA OF ∠DEF OE The center of a regular polygon is the common center for the inscribed and circum- scribed circles of the polygon. DEFINITION NOTE: The preceding definition does not tell us how to locate the center of a regular polygon. The center is the intersection of the angle bisectors of two consecutive angles; alternatively, the intersection of the perpendicular bisectors of two consecutive sides can be used to locate the center of the regular polygon. Note that a regular polygon has a center, whether or not either of the related circles is shown. In Figure 7.30, point O is the center of the regular pentagon RSTVW. In this figure, is called a “radius” of the regular pentagon. OR A radius of a regular polygon is any line segment that joins the center of the regular polygon to one of its vertices. DEFINITION V W R S T O Figure 7.30 쮿 A B C F E D Q Figure 7.32 In the proof of Theorem 7.3.1, we saw that “All radii of a regular polygon are congruent.” R Y X U V W S T Q P Figure 7.31 An apothem of a regular polygon is any line segment drawn from the center of that polygon perpendicular to one of the sides. DEFINITION In regular octagon RSTUVWXY with center P see Figure 7.31, the segment is an apothem. Any regular polygon of n sides has n apothems and n radii. The proof of Theorem 7.3.1 establishes that “All apothems of a regular polygon are congruent.” PQ A central angle of a regular polygon is an angle formed by two consecutive radii of the regular polygon. DEFINITION In regular hexagon ABCDEF with center Q see Figure 7.32, angle EQD is a cen- tral angle. Due to the congruences of the triangles in the proof of Theorem 7.3.1, we see that “All central angles of a regular polygon are congruent.” This leads to Theorem 7.3.2. The measure of the central angle of a regular polygon of n sides is given by c = 360 n . THEOREM 7.3.2 We apply Theorem 7.3.2 in Example 4. EXAMPLE 4 a Find the measure of the central angle of a regular polygon of 9 sides. b Find the number of sides of a regular polygon whose central angle measures 72°. Solution a b 72 = 360 n : 72n = 360 : n = 5 sides c = 360 9 = 40° 쮿 The next two theorems follow from the proof of Theorem 7.3.1. Any radius of a regular polygon bisects the angle at the vertex to which it is drawn. THEOREM 7.3.3 Any apothem of a regular polygon bisects the side of the polygon to which it is drawn. THEOREM 7.3.4