Math Section 4.4 Equidistance Theorems By: Dan Schlosser and Brad Wagner.

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Presentation transcript:

Math Section 4.4 Equidistance Theorems By: Dan Schlosser and Brad Wagner

If 2 points are the same distance from a 3 rd point they are said to be equidistant from that point. AC B AB is equidistant to CB.

Look at these 2 figures. AND A D N S B R A D W W

In both cases the white line is the perpendicular bisector of the black line. The definition of a perpendicular bisector is a line that bisects and is perpendicular to another line.

Theorems that deal with perpendicular bisectors are: Theorem 24: if 2 points are each equidistant from the end-points of a segment, then the 2 points determine the perpendicular bisector of that segment. Theorem 25: if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoint of that segment.

Sample problem using theorem 24 A BD E Given: AB AD= ~ AE BD Prove: AE is the perpendicular bisector of BD EAD EAB= ~

Answer: StatementReason 1. AB AD= ~ 1. Given 2. AE BD 2. Given 3. AED and AEB are right s = ~ 4. AED AEB 5. AE AE = = ~ 6. EAD EAB ~ = 7. EAD EAB= ~ 8. BE DE= ~ 9. AE bisector of BD 3. lines form right s 4. right are= ~ 5. reflexive 6. Given 7. ASA (4, 5, 6)‏ 8. CPCTC 9. if 2 points are each equidistant from the end- points of a segment, then the 2 points determine the perpendicular bisector of that segment.

Sample problem using theorem 25 W XY Z Given: WZ is the bisector of XY. Prove: WXZ WYZ= ~

Answer: StatementReason 1. WZ bisector of XY1. Given 2. WX WY= ~ 2. if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoint of that segment. 3. XZ YZ= ~ 3. same as 24. WZ WZ= ~ 4. reflexive5. WXZ WYZ= ~ 5. SSS (2, 3, 4)‏6. WXZ WYZ= ~ 6. CPCTC

Practice problem A BC D Given: AD is the bisector of XY and Find: the perimeter of ABC AC = 12, BD = 5

Answer: If BD=5 then to get the length of BC you must do 5x2 to get BC=10. Then since AC=12, AB=12 because AD is the bisector of BC. Therefore the perimeter of ABC is 34 because =24.

Practice problem MN O P Given: PO is the bisector of MN and PNO = 70  Find: m MPN

Answer: If PNO=70  then PMO=70  because PO is perpendicular bisector of MN. 70  + 70  = 140  and since we know a triangles angles must equal 180 , 180  – 140  = 40 . Which means MPN = 40 .

Works Cited Milauskas, George, and Robert Whipple. Geometry for Enjoyment and Challenge. Boston: Houghton Mifflin Company 1991.Print.