Verify this locus theorem:

The point at which the perpendicular bisectors of the sides of a triangle meet is the circumcenter of the triangle. The term circumcenter is easily remembered as the cen- ter of the circumscribed circle. B A C E N P M Figure 7.15 EXAMPLE 4 Complete the construction of the circumscribed circle for that was given in Figure 7.16a. 䉭ABC EXAMPLE 2 Complete the construction of the inscribed circle for in Figure 7.14b. Solution Having found the incenter E, we need the length of the radius. Because as shown in Figure 7.15, the length of or or is the desired radius; thus, the circle is completed. NOTE: The sides of the triangle are tangents for the inscribed circle or incircle of the triangle. The incircle lies inside the triangle. EP EM EN EN ⬜ AC 䉭ABC 쮿 It is also possible to circumscribe a circle about a given triangle. The construction depends on the following theorem, the proof of which is sketched in Example 3. Reminder A point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment. The three perpendicular bisectors of the sides of a triangle are concurrent. THEOREM 7.2.2 EXAMPLE 3 Give an informal proof of Theorem 7.2.2. See in Figure 7.16. Proof Let and name the perpendicular bisectors of sides and , respectively. See Figure 7.16a. Using Theorem 7.1.2, the point of concurrency F is equidistant from the endpoints of ; thus, . In the same manner, . By the Transitive Property, it follows that ; again citing Theorem 7.1.2, F must be on the perpendicular bisector of because this point is equidistant from the endpoints of . Thus, F is the point of concurrence. AB AB AF ⬵ BF AF ⬵ FC BF ⬵ FC BC AC BC FR FS 䉭ABC Exs. 3–7 a B A C S F R Figure 7.16 b B A C S F R 쮿 The incenter and the circumcenter of a triangle are generally distinct points. How- ever, it is possible for the two centers to coincide in a special type of triangle. Although the incenter of a triangle always lies in the interior of the triangle, the circumcenter of an obtuse triangle will lie in the exterior of the triangle. See Figure 7.17. The circum- center of a right triangle is the midpoint of the hypotenuse. To complete the discussion of concurrence, we include a theorem involving the altitudes of a triangle and a theorem involving the medians of a triangle. Solution We have already identified the center of the circle as point F. To complete the construction, we use F as the center and a radius of length equal to the distance from F to any one of the vertices A, B, or C. The circumscribed circle is shown in Figure 7.16b. NOTE: The sides of the inscribed triangle are chords of the circumscribed circle, which is called the circumcircle of the triangle. The circumcircle of a polygon lies outside the polygon except where it contains the vertices of the polygon. 쮿 Figure 7.17 Exs. 8–12 The three altitudes of a triangle are concurrent. THEOREM 7.2.3 E a D F N Figure 7.18 b S R X T The point of concurrence for the three altitudes of a triangle is the orthocenter of the triangle. In Figure 7.18a, point N is the orthocenter of . For the obtuse tri- angle in Figure 7.18b, we see that orthocenter X lies in the exterior of . Rather than proving Theorem 7.2.3, we sketch a part of that proof. In Figure 7.19a, is shown with its altitudes. To prove that the altitudes are concurrent requires

1. that we draw auxiliary lines through N parallel to , through M parallel to

, and through P parallel to . [See Figure 7.19b.]

2. that we show that the altitudes of are perpendicular bisectors of the

sides of the newly formed ; thus altitudes , , and are concurrent a consequence of Theorem 7.2.2. SKETCH OF PROOF THAT IS THE BISECTOR OF : Because is an altitude of , . But by construction. Because a line perpendicular to one of two parallel lines must be perpendicular to the other, we have . Now we need to show that bisects . By construction, and , so MRPN is a parallelogram. Then because the opposite sides of a parallelogram are congruent. By construction, MPSN is also a parallelogram and . By the Transitive Property of Congruence, . Thus, is bisected at point P, and is the bisector of . RS ⬜ PX RS RP ⬵ PS MN ⬵ PS MN ⬵ RP RP 7 MN MR 7 NP RS PX PX ⬜ RS RS 7 MN PX ⬜ MN 䉭MNP PX RS ⬜ NZ MY PX 䉭RST 䉭MNP NM NP MP 䉭MNP 䉭RST 䉭DEF Z M P N Y X a Figure 7.19 b Y Z M P T S R N X PX In Figure 7.19b, similar arguments leading to one long proof could be used to show that is the bisector of and also that is the bisector of . Because the concurrent perpendicular bisectors of the sides of are also the altitudes of , these altitudes must be concurrent. The intersection of any two altitudes determines the orthocenter of a triangle. We use this fact in Example 5. If the third altitude were constructed, it would contain the same point of intersection the orthocenter. 䉭MNP 䉭RST TR ⬜ MY TS ⬜ NZ EXAMPLE 5 Construct the orthocenter of in Figure 7.20. Solution First construct the altitude from A to ; here, we draw an arc from A to intersect at X and Y. Now draw equal arcs from X and Y to intersect at Z. is the desired altitude. Repeat the process to construct altitude from vertex C to side . The point of intersection O is the orthocenter of . 䉭ABC AB CJ AH BC BC 䉭ABC Recall that a median of a triangle joins a vertex to the midpoint of the opposite side of the triangle. Through construction, we can show that the three medians of a triangle are concurrent. We will discuss the proof of the following theorem in Chapter 10. 쮿 B O C Z Y X H A J Figure 7.20 Exs. 13–16 The three medians of a triangle are concurrent at a point that is two-thirds the distance from any vertex to the midpoint of the opposite side. THEOREM 7.2.4 The point of concurrence for the three medians is the centroid of the triangle. In Figure 7.21, point C is the centroid of . According to Theorem 7.2.4, , and TC = 2 3 TP. SC = 2 3 SN, RC = 2 3 RM 䉭RST Discover On a piece of paper, draw a triangle and its medians. Label the figure the same as Figure 7.21. a Find the value of . b Find the value of . ANSWERS SC CN RC RM a b or 2 2 1 2 3 S C R T N M P Figure 7.21 EXAMPLE 6 Suppose that the medians of in Figure 7.21 have the lengths , , and . The centroid of is point C. Find the length of: a RC b CM c SC Solution a , so . b , so . c , so . SC = 2 3 15 = 10 SC = 2 3 SN CM = 12 - 8 = 4 CM = RM - RC RC = 2 3 12 = 8 RC = 2 3 RM 䉭RST TP = 18 SN = 15 RM = 12 䉭RST 쮿 The following relationships are also implied by Theorem 7.2.4. In Figure 7.21, , , and . Equivalently, , , and . PC = 1 2 CT CN = 1 2 SC CM = 1 2 RC CT = 2PC SC = 2CN RC = 2 CM Solution Median separates into two congruent right triangles, and ; this follows from SSS. With Z the midpoint of , . Using the Pythagorean Theorem with in Figure 7.22b, we have By Theorem 7.2.4, Because it follows that QZ = 4. QZ = 1 2 RQ, RQ = 2 3 RZ = 2 3 12 = 8 RZ = 12 RZ 2 = 144 225 = RZ 2 + 81 15 2 = RZ 2 + 9 2 RS 2 = RZ 2 + SZ 2 䉭RZS SZ = 9 ST 䉭RZT 䉭RZS 䉭RST RZ 쮿 EXAMPLE 7 GIVEN: In Figure 7.22a, isosceles with , and ; medians , , and meet at centroid Q. FIND: RQ and QZ SY TX RZ ST = 18 RS = RT = 15 䉭RST R S T Y Q X Z a Figure 7.22 R S Q Z b 15 9 Exs. 17–22 It is possible for the angle bisectors of certain quadrilaterals to be concurrent. Likewise, the perpendicular bisectors of the sides of a quadrilateral can be concurrent. Of course, there are four angle bisectors and four perpendicular bisectors of sides to consider. In Example 8, we explore this situation.