Geometry Lesson 5 – 1 Bisectors of Triangles Objective: Identify and use perpendicular bisectors in triangles. Identify and use angle bisectors in triangles.

Perpendicular Bisector Perpendicular bisector Any segment, line, or plane that intersects a segment at its midpoint forming a right angle.

Theorems Perpendicular Bisector Theorem If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. Converse of Perpendicular Bisector Theorem If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

Find AB AB = 4.1

Find WY WY = 3

Find RT RQ = RT 2x + 3 = 4x – 7 3 = 2x – 7 10 = 2x 5 = x RT = 4x – 7 = 4(5) – 7 = 20 – 7 = 13

Concurrent lines 3 or more lines intersect at a point Point of Concurrency The point where 3 or more lines intersect.

Perpendicular Bisectors Acute: Interior Right: On the Triangle Obtuse: Exterior

Circumcenter The point of concurrency of the perpendicular bisectors Circumcenter Theorem The perpendicular bisectors of a triangle intersect at a point called the circumcenter that is equidistant from the vertices of the triangle.

Angle Bisector Angle bisector A line, segment, or ray that cuts an angle into 2 congruent parts.

Theorem Angle Bisector Theorem If a point is on the bisector of an angle, then it is equidistant from the sides of that angle. Converse Angle Bisector Theorem If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle.

Find XY XY = 7

Find the measure of angle JKL 37

Find SP 3x + 5 = 6x – 7 5 = 3x – 7 12 = 3x 4 = x SP = 6x – 7 = 6(4) – 7 = 17 SP = 17

Angle bisectors of a triangle Notice all angle bisectors go through a vertex and Intersect in the interior of the triangle.

Incenter The point of concurrency of the angle bisectors of a triangle. Incenter Theorem The angle bisectors of a triangle intersect at a point called the incenter that is equidistant from each side of the triangle.

Find each measure if J is the incenter. JF JF = JE How can we find JE? (JE) = 15 2 (JE) = 225 (JE) 2 = 81 JE = 9 JF = 9

If P is the incenter find the following. PK PK = PJ (PJ) = 20 2 (PJ) = 400 (PJ) 2 = 256 PJ = 16 PK =

Homework Pg – 8 all, 10 – 14 E, 18 – 34 E, 60 – 64 E