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# 5.1 perpendiculars and bisectors i

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### 5.1 perpendiculars and bisectors i

1. 1. Perpendiculars and Bisectors I
2. 2. Perpendicular Bisector <ul><li>A line, ray, segment or plane perpendicular to a segment at its midpoint. </li></ul>is a ⊥ bisector of
3. 4. Constructions <ul><li>Construct a perpendicular bisector to a line segment . </li></ul><ul><li>Construct a perpendicular to a line l , through a point P on l . </li></ul>
4. 5. Equidistant <ul><li>A point is equidistant from two points if its distance from each point is the same . </li></ul><ul><li>C is equidistant from A and B , since C was drawn so that CA = CB . </li></ul>
5. 7. Theorem: Perpendicular Bisector <ul><li>If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. </li></ul>
6. 8. Proof of the Perpendicular Bisector Theorem Given: Prove: Show ∆ ACP ≅ ∆ BCP CA = CB ∠ APC ≅ ∠ CPB Reflexive property of congruence ∆ ACP ≅ ∆ BCP SAS Postulate CPCTC CA = CB Definition of Congruence Plan: bisects bisects
7. 9. Theorem: Perpendicular Bisector Converse <ul><li>If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment. </li></ul>If DA = DB , then D lies on the perpendicular bisector of
8. 10. Prove the Perpendicular Bisector Theorem Converse Given : C is equidistant from A and B Prove : C is on the perpendicular bisector of Plan Show ∆ APC ≅ ∆ BPC Draw
9. 11. Example
10. 12. Homework <ul><li>Exercise 5.1 page 267: 1-10, 14, 16-18, 29. </li></ul>