1 / 4

EXAMPLE 1

Write a two-column proof for the situation in Example 4 on page 107 . m ∠ 1 = m ∠ 3. GIVEN:. m ∠ EBA = m ∠ DBC. PROVE:. REASONS. STATEMENT . 1. 1. m ∠ 1 = m ∠ 3. Given. 2. Angle Addition Postulate. 2. m ∠ EBA = m ∠ 3 + m ∠ 2. 3.

talasi
Télécharger la présentation

EXAMPLE 1

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. Write a two-column proof for the situation in Example 4 on page 107. m∠ 1=m∠ 3 GIVEN: m∠ EBA= m∠ DBC PROVE: REASONS STATEMENT 1. 1. m∠ 1=m∠ 3 Given 2. Angle Addition Postulate 2. m∠ EBA =m∠ 3 + m∠ 2 3. Substitution Property of Equality 3. m∠ EBA=m∠ 1 + m∠ 2 EXAMPLE 1 Write a two-column proof

  2. 4. Angle Addition Postulate m∠ EBA= m∠ DBC 5. 5. Transitive Property of Equality EXAMPLE 1 Write a two-column proof 4. m∠ 1 +m∠ 2 = m∠ DBC

  3. 1. Four steps of a proof are shown. Give the reasons for the last two steps. REASONS STATEMENT 1. 1. AC = AB + AB Given 2. 2. AB + BC = AC Segment Addition Postulate ? 3. 3. AB + AB = AB + BC ? 4. 4. AB = BC for Example 1 GUIDED PRACTICE GIVEN :AC = AB + AB PROVE :AB = BC

  4. ANSWER GIVEN :AC = AB + AB PROVE :AB = BC REASONS STATEMENT 1. 1. AC = AB + AB Given 2. 2. AB + BC = AC Segment Addition Postulate 3. 3. AB + AB = AB + BC Transitive Property of Equality 4. 4. AB = BC Subtraction Property of Equality for Example 1 GUIDED PRACTICE

More Related