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

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

8. Given: Prove: S S S 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) S
3. PBT (1) S S

9. TRUE/FALSE PRACTICE Ready??

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

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

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

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

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

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

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

17. T R U E OR F A L S
???
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. P 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
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. E 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 B C 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)