The EQUIDISTANCE Theorems

1. The EQUIDISTANCE Theorems Advanced Geometry Chapter 4.4

2. Some basics: Definition A.) Distance – the length of the path between two objects. B.) A line segment is the shortestpath between two points. Postulate B A

3. What does Equidistance look like? 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 X

4. Definition of Perpendicular Bisector: A perpendicular bisector of a segment is the line that both _________ and is _____________ to the segment. BISECTS PERPENDICULAR

5. Theorem 24:If 2 points are equidistant from the endpoints of a segment, then the two points DETERMINE the PERPENDICULAR BISECTOR of the segment. (Perpendicular Bisector Theorem) NEEDED: 2 points equidistant from endpoints of same segment, or 2 pairs of congruent segments

6. D Given: ΔABC ≅ ΔADC C C A A E Prove: AC bis of BD B 1) Given 1) ΔABC ≅ΔADC A A 2) AB ≅ AD 2) CPCTC A • If 2≅segs, then a point is =dist • from endpts of another seg 3) A is =dist from B & D C C 4) CB ≅ CD 4) CPCTC C 5) C is =dist from B & D 5) Same as #3 6) PBT (Perpendicular Bisector Theorem) • If 2 points are =dist from • endpts of same seg, then they determine the perpendicular bisector of the seg. (3, 5) 6) AC is  bis of BD

7. Theorem 25:If a point is ONthe perpendicular bisector of a segment, then . . . Given:HI is  bisector of LO I A B C L O H Prove: IL  IO Prove: BL  BO Prove: AL  AO Prove: HL  HO . . . the point is EQUIDISTANT from the endpoints of the segment. (Perpendicular Bisector Theorem) NEEDED: Perpendicular Bisector of the Segment.

8. 1. 2. 3. X is =dist from A & B 4. 5. F is =dist from A & B 6. 7. 1. Given 2. Reflexive 3.If a point is on the perp bisector, then it is =dist. (1) 4. A bis  seg into 2 ≅ segs (3) 5. Same as #3 (1) 6. Same as #4 (5) 7. SSS (2,4,6) Given:Prove: F X S 3. PBT (1) S S

9. TRUE/FALSEPRACTICEReady??

10. T R U E OR F A L S E ??? Given: 1. E is the midpoint of BC. TRUE

11. T R U E OR F A L S E ??? Given: 2. ∡AEC is a right angle TRUE

12. T R U E OR F A L S E ??? Given: 3. E is the midpoint of AD FALSE

13. T R U E OR F A L S E ??? Given: 4. AC ≅ AB TRUE

14. T R U E OR F A L S E ??? Given: 5. CE ≅ BE TRUE

15. T R U E OR F A L S E ??? Given: 6. CA ≅ CD FALSE

16. T R U E OR F A L S E ??? Given: 7. AE ≅ ED FALSE

17. T R U E OR F A L S E ??? Given: 8. CB bisects AD FALSE

18. Prove the following statement: The line drawn from the vertex angle of an isosceles triangle through the point of intersection of the medians to the legs is perpendicular to the base.

19. A Given: ΔABC is isosceles, base BC BF is median to side AC CE is median to side AB Prove: AD is the altitude to BC E Since ABC is isosceles with base BC, AB and AC are congruent legs. This also gives us congruent base angles, ∡ABC and ∡ACB. BF and CE are medians to the legs, AC and AB, respectively. This means that EB and FC are congruent by division property. BC is a reflexive side, so Δ EBC and ΔFCB are congruent by SAS. We now have proven ∡PBC and ∡PCB congruent by CPCTC. Now, ΔBPC is isosceles, with base BC and congruent legs PB and PC. F Or, we could say now that because of our isosceles triangles giving us 2 pairs of are congruent segs, two points (A and P) are equidistant from the endpoints of segment (BC), so they determine the PERPENDICULAR bisector of the segment by PBT. Now, since D is ON line AP, by relying upon perpendicular bisector theorem again we can conclude that AD is the altitude to BC! P C B D D Now with PB = PC, and we know that PD is a reflexive side of congruent Δ’s DPB and DPC (∡PDB and ∡PDC are supplementary and adjacent, which means right angles. Right ∡’s ΔDPB  ΔDPC by HL) Then, Right ∡s   segs, ∴ AD is the altitude to BC !

20. Which method do you prefer???

21. 4.4 Homework Pp 187 – 190 (2 – 4; 7 – 10; 12, 15, 18, 19)