1 / 9

Perpendicular and angle bisectors

Perpendicular and angle bisectors. Geometry (Holt 5-1) K. Santos. Perpendicular Bisector Theorem (and converse). If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. X Given: is the perpendicular

dori
Download Presentation

Perpendicular and angle bisectors

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Perpendicular and angle bisectors Geometry (Holt 5-1) K. Santos

  2. Perpendicular Bisector Theorem (and converse) If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. X Given: is the perpendicular bisector of A B Then: XA = XB Y

  3. Example—Perpendicular Bisector Find MN. N 2.6 M L MN = NL since NL = 2.6 then MN = 2.6

  4. Example—Perpendicular Bisector B Find BC. 38 D A 12 38 C BD = DC So BD = 12 BC = BD + DC BC = 12 + 12 BC =24

  5. Example—Perpendicular Bisector U Find TU 3x + 9 7x - 15 T W V UV = TU 7x -15 = 3x + 9 4x – 15 = 9 4x = 24 x = 6 TU = 3x + 9 TU = 3(6) + 9 = 27

  6. Angle Bisector Theorem (and converse) If a point is on the bisector of angle, then it is equidistant from the sides of the angle. Given: bisects <APB A P C Then: AC = BC B

  7. Example—Angle Bisector Find BC B CD = 7.2 A C D BC = CD Since CD = 7.2 then BC = 7.2

  8. Example—Angle Bisector F Find m< EFH. m< EFG = 50 E G H m< EFG = 50 m < EFH = ½ m<EFG m < EFH = ½(50) = 25

  9. Example—Angle Bisector K 2a + 26 Find m <MKL. L 3a +20 J M m < JKM = m <MKL 3a + 20 = 2a + 26 a + 20 = 26 a = 6 m< MKL = 2a + 26 m< MKL = 2(6) + 26 = 38

More Related