1 / 7

Proving Segment Relationships

Proving Segment Relationships. Postulate 2.8 - The Ruler Postulate The points on any line or line segment can be paired with real numbers so that, given any two points A and B on a line, A corresponds to zero and B corresponds to a positive real number. Proving Segment Relationships.

makoto
Télécharger la présentation

Proving Segment Relationships

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. Proving Segment Relationships Postulate 2.8 - The Ruler Postulate • The points on any line or line segment can be paired with real numbers so that, given any two points A and B on a line, A corresponds to zero and B corresponds to a positive real number.

  2. Proving Segment Relationships Postulate 2.9 – Segment Addition Postulate Given three collinear points A, B, and C, if B is between A and C, then AB + BC = AC. Likewise, if AB + BC = AC, then B is between A and C.

  3. Proving Segment Relationships Theorem 2.2 – Segment Congruence Theorem Congruence of segments is reflexive, symmetric, and transitive. Reflexive: segment AB  segment AB Symmetric: If segment AB  segment CD, then segment CD  segment AB. Transitive: If segment AB  segment CD and segment CD  segment EF, then segment AB  segment EF.

  4. Prove the following. Given: Prove: Example 7-1b

  5. Proof: Statements Reasons 1. 1. Given AC = AB, AB = BX 2. 2. Transitive Property AC = BX CY = XD 3. 3. Given 4. AC + CY = BX + XD 4. Addition Property 5. 5. Segment Addition Property AC + CY = AY; BX + XD = BD 6. 6. Substitution AY = BD Example 7-1c

  6. Prove the following. Given: Prove: Example 7-2c

  7. Statements Reasons 1. 1. Given 2. 2. Transitive Property 3. 3. Given 4. 4. Transitive Property 5. 5. Symmetric Property Example 7-2d Proof:

More Related