1 / 13

GBK Geometry

GBK Geometry. Jordan Johnson. Today’s plan. Greeting HW Questions? Take out log / HW Inequality Review Theorems & Proofs: Perpendicular Bisectors Homework / Questions Clean-up. Inequalities. Solve: (2 x + 3) ⁄ 5 ≥ 7 –( x – 4) < 2 –10 x + 3 ≤ 23 Question to think about:

obelia
Download Presentation

GBK Geometry

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. GBK Geometry Jordan Johnson

  2. Today’s plan • Greeting • HW Questions? • Take out log / HW • Inequality Review • Theorems & Proofs: Perpendicular Bisectors • Homework / Questions • Clean-up

  3. Inequalities • Solve: • (2x + 3)⁄5≥ 7 • –(x – 4) < 2 • –10x + 3 ≤23 • Question to think about: • What’s the smallest number of sides a polygon can have if it has three collinear vertices? • Note: It can’t be degenerate, and it can’t cross itself.

  4. Proof: Ch. 6 Lesson 1 • Theorem 16: • In a plane, two points that are each equidistant from the endpoints of a line segment determine the perpendicular bisector of that line segment. • Given: C and D are each equidistant from A and B. • Prove: AB  DC and DC bisects AB. • Two possibilities: • C or D is on AB • Neither is on AB trivial really easy

  5. Proof of Theorem 16 • Case I: C is on AB. C is equidistant from A and B, soAC = BC, and C is the midpoint of AB by definition.Therefore, CD bisects AB. Since AC = BC, AD = BD, and (by the reflexive property)CD = CD, we have ΔACD ΔBCD by SSS. By congruence, 1 = 2; because 1 and 2 form a linear pair and are equal, they are right angles, and so CD  AB. Since CD bisects AB and CD  AB,CD is the perpendicular bisector of AB, QED. 2 1

  6. Proof: Ch. 6 Lesson 1 • Case II: Neither C nor D is on AB. • Given: C and D are each equidistant from A and B. • Prove: AB  DC and DC bisects AB. • First step: Let E be the intersection of AB and CD.

  7. Converse to Theorem 16 • Theorem: • Every point on the perpendicular bisector of segment AB is equidistant from A and B. • Rewrite as a conditional statement. • One way: • If a point P lies on the perpendicular bisector of AB, P is equidistant from A and B. • Proof: • Given: Point P on the perpendicular bisector of AB. • Prove: PA = PB. • 2 cases: P lies on AB, P doesn’t lie on AB.

  8. Homework • 25 minutes • Study/write proofs • Theorem 16 and its converse • Asg #43 • Read Ch. 6 Lesson 1 for definition of symmetry, too. • Khan Academy • Practice solving inequalities

  9. Clean-up / Reminders • Pick up all trash / items. • Push in chairs (at front and back tables). • See you tomorrow!

  10. Unused slides follow

  11. Defining Reflection Symmetry • Remember reflection symmetry: • A figure has reflection symmetry about a line m iff… …it is its own image when reflected about line m.

  12. Alternate definition • Points A and B are symmetric with respect to line m iff m is the perpendicular bisector of AB.

  13. Alternate definition, cont. • Two figures a and b are symmetric w.r.t. line m iff every point in a is symmetric with some point in b (w.r.t. line m). • (A figure has reflection symmetry w.r.t. line m iff it is symmetric with itself w.r.t line m.) a b

More Related