1. AD bisects BAC. Find x. ANSWER 20 2. Solve x² + 12² = 169. ANSWER

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1. AD bisects BAC. Find x. ANSWER 20 2. Solve x² + 12² = 169. ANSWER ±5 3. The legs of a right triangle measure 15 feet and 20 feet. Find the length of the hypotenuse. ANSWER 25 ft

Use properties of special segments in triangles. Target Use properties of special segments in triangles. You will… Use perpendicular bisectors to solve problems.

Vocabulary Angle Bisector Theorem 5.5 – If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle. Angle Bisector Converse Theorem 5.6 – If a point lies in the interior of an angle and is equidistant from the sides of the angle, then it lies on the angle bisector.

Vocabulary Concurrency of Angle Bisectors of a Triangle Theorem 5.7 – The angle bisectors of a triangle intersect at a point (incenter) that is equidistant from the sides of the triangle. P is the incenter of triangle ABC The incenter of a triangle always lies in the interior of the triangle.

EXAMPLE 1 Use the Angle Bisector Theorems Find the measure of GFJ. SOLUTION Because JG FG and JH FH and JG = JH = 7, FJ bisects GFH by the Converse of the Angle Bisector Theorem. So, mGFJ = mHFJ = 42°.

EXAMPLE 2 Solve a real-world problem SOCCER A soccer goalie’s position relative to the ball and goalposts forms congruent angles, as shown. Will the goalie have to move farther to block a shot toward the right goalpost R or the left goalpost L?

EXAMPLE 2 Solve a real-world problem SOLUTION The congruent angles tell you that the goalie is on the bisector of LBR. By the Angle Bisector Theorem, the goalie is equidistant from BR and BL . So, the goalie must move the same distance to block either shot.

Use algebra to solve a problem EXAMPLE 3 Use algebra to solve a problem ALGEBRA For what value of x does P lie on the bisector of A? SOLUTION From the Converse of the Angle Bisector Theorem, you know that P lies on the bisector of A if P is equidistant from the sides of A, so when BP = CP. BP = CP Set segment lengths equal. x + 3 = 2x –1 Substitute expressions for segment lengths. 4 = x Solve for x. Point P lies on the bisector of A when x = 4.

GUIDED PRACTICE for Examples 1, 2, and 3 In Exercises 1–3, find the value of x. 1. 2. 3. 15 ANSWER 11 ANSWER 5 ANSWER

GUIDED PRACTICE for Examples 1, 2, and 3 4. Do you have enough information to conclude that QS bisects PQR? Explain. No; you need to establish that SR QR and SP QP. ANSWER

EXAMPLE 4 Use the concurrency of angle bisectors In the diagram, N is the incenter of ABC. Find ND. SOLUTION By the Concurrency of Angle Bisectors of a Triangle Theorem, the incenter N is equidistant from the sides of ABC. So, to find ND, you can find NF in NAF. Use the Pythagorean Theorem stated on page 18.

Use the concurrency of angle bisectors EXAMPLE 4 Use the concurrency of angle bisectors c = 2 a + b Pythagorean Theorem NF + 16 2 20 = Substitute known values. 400 = NF + 256 2 Multiply. 144 = NF 2 Subtract 256 from each side. 12 = NF Take the positive square root of each side. Because NF = ND, ND = 12.

GUIDED PRACTICE for Example 4 5. WHAT IF? In Example 4, suppose you are not given AF or AN, but you are given that BF = 12 and BN = 13. Find ND. 5 ANSWER