The EQUIDISTANCE Theorems Advanced Geometry Chapter 4.4.
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Transcript of The EQUIDISTANCE Theorems Advanced Geometry Chapter 4.4.
The EQUIDISTANCE The EQUIDISTANCE TheoremsTheorems
The EQUIDISTANCE The EQUIDISTANCE TheoremsTheorems
Advanced Geometry Chapter 4.4Advanced Geometry Chapter 4.4
Some basics:
A.) Distance – the length of the path between two objects.
B.) A line segment is the shortest path between two points.
segment itselfAB
length of segmentAB
Definition
Postulate
A
B
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
Q
P
XX
Definition of Perpendicular Bisector:
A perpendicular bisector of a segment is the line that both _________ and is _____________ to the segment.
BISECTSPERPENDICUL
AR
A B
L
X
is bisector ofL AB
Theorem 24: If 2 points are equidistant from the endpoints of a segment,
NEEDED: 2 points equidistantfrom endpoints of same
segment, or
then the two points DETERMINE the PERPENDICULAR BISECTOR of the segment. (Perpendicular Bisector Theorem)
2 pairs of congruent segments
A C
6) PBT (Perpendicular Bisector Theorem)
AA AA
CCCC
Given: ΔABC ≅ ΔADC
Prove: AC bis of BDA E
C
D
B
1) ΔABC ≅ ΔADC 1) Given
2) AB ≅ AD 2) CPCTC
AA3) A is =dist from B & D 3) If 2≅segs, then a point is =dist
from endpts of another seg4) CB ≅ CD
CC
4) CPCTC
5) C is =dist from B & D 5) Same as #3
6) AC is bis of BD 6) If 2 points are =dist from endpts of same seg, then
they determine the perpendicular bisector of the seg. (3, 5)
Prove: AL AO
Theorem 25: If a point is ON the perpendicular bisector of a segment, then . . .
NEEDED: Perpendicular Bisector of the Segment.
. . . the point is EQUIDISTANT from the endpoints of the segment. (Perpendicular Bisector Theorem)
OL
Given: HI is bisector of LO
C
B
AI
H
Prove: IL IO Prove: BL BO
Prove: HL HO
Given:Prove:
1.
2.
3. X is =dist from A & B
4. 5. F is =dist from A & B6. 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)
Bisector ofEF ABBFXAFX
A
FX
B
E
A
FX
B
E
ABofBisEF
XFXF
XBXA
FBFA BFXAFX
S
S
S
3. PBT (1)
FX
TRUE/FALSEPRACTICE
Ready??
C
E
B
DA
BCoftorbiAD sec
TRUE
TRUE
OR
FALSE
???
Given:
1. E is the midpoint of BC.
C
E
B
DA
BCoftorbiAD sec
2. ∡AEC is a right angle
TRUE
Given:
TRUE
OR
FALSE
???
C
E
B
DA
BCoftorbiAD sec
FALSE
3. E is the midpoint of AD
Given:
TRUE
OR
FALSE
???
C
E
B
DA
BCoftorbiAD sec
TRUE
4. AC ≅ AB
Given:
TRUE
OR
FALSE
???
C
E
B
DA
BCoftorbiAD sec
TRUE
5. CE ≅ BE
Given:
TRUE
OR
FALSE
???
C
E
B
DA
BCoftorbiAD sec
FALSE
6. CA ≅ CD
Given:
TRUE
OR
FALSE
???
C
E
B
DA
BCoftorbiAD sec
FALSE
7. AE ≅ ED
Given:
TRUE
OR
FALSE
???
C
E
B
DA
BCoftorbiAD sec
FALSE
8. CB bisects AD
Given:
TRUE
OR
FALSE
???
ProveProve 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.
Given: Given: ΔΔABC is isosceles, base BCABC is isosceles, base BC BF is median to side ACBF is median to side AC
CE is median to side AB CE is median to side AB Prove: AD is the altitude to BCProve: AD is the altitude to BC
DD
AA
CCBB
EE FFSince 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.
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 !
PPOr, 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!
D
Which method do you
prefer???
4.4 Homework4.4 Homework
Pp 187 – 190
(2 – 4; 7 – 10; 12, 15, 18, 19)