A Right Angle Theorem and the Equidistance Theorems

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A Right Angle Theorem and the Equidistance Theorems. Advanced Geometry 4.3 and 4.4. You did it! Add to your Index Cards!. Right Angle Theorem: If two angles are both supplementary and congruent, then they are right angles. Example. Given: Circle P S is the midpoint - PowerPoint PPT Presentation

Transcript of A Right Angle Theorem and the Equidistance Theorems

EQUIDISTANCE

Advanced Geometry 4.3 and 4.4A Right Angle Theorem and the Equidistance Theorems1You did it! Add to your Index Cards!Right Angle Theorem:

If two angles are both supplementaryand congruent, then they are rightangles.

Example

Given: Circle P S is the midpoint of

Prove:

3Example

Given: Circle P S is the midpoint of Prove:

StatementsReasons1. Circle P1. Given2. S is the midpoint of 2. Given3. 3. Def of midpt4. 4. Reflexive5. Draw and 5. Two points det a seg6. 6. All radii of a circle are

SSS8. 8. CPCTC9. are supp9. Def of supp10. are rt 10. If two angles are supp and congruent, then they are right angles.11. 11. Def of perpendicular

4A.) (def) Distance (between two objects) is the length of the shortest path joining them.

B.) (postulate) A line segment is the shortest path between two points.

Index CardAB5Definition of Equidistance:

If 2 points P and Q are the same distance from a third point X, then X is said to be _____________from P and Q.EQUIDISTANT

Another index card6Definition of a Perpendicular Bisector:A perpendicular bisector of a segment is the line that both _________ and is _____________to the segment.BISECTSPERPENDICULAR

Index Card7

Theorem: If 2 points are equidistant from the endpoints of a segment, then they determine the perpendicular bisector of that segment.

NEEDED: 2 points equidistant or 2 pairs congruent segmentsP and Q are two points that are equidistant from E and D (the endpoints of segment ED), so they determine the perpendicular bisector of the segment (ED).EDQPIndex Card8Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

NEEDED: Perpendicular bisector of the segment EDQPIf N is on PQ, the perpendicular bisector of ED, then N is equidistant from E and D. (EN = ND)NIndex Card9TRUE/FALSEPRACTICE

Ready??10

1. E is the midpoint of BC.TRUE11

2.


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