Transcript File

A Right Angle Theorem and the
Equidistance Theorems
Advanced Geometry 4.3 and 4.4
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Right Angle Theorem:
If two angles are both supplementary
and congruent, then they are right
angles.
Example
 Given: Circle P
S is the midpoint
of
 Prove:
P
Q
S
R

Given: Circle P
S is the midpoint of
Example
Prove:
Statements
Reasons
1. Circle P
1. Given
2. S is the midpoint of
Q
S
R
2. Given
3.
QS  SR
3. Def of midpt
4.
PS  PS
4. Reflexive
5. Draw PQ and
P
QR
5. Two points det a seg
PR
6.
PQ  PR
6. All radii of a circle are 
7.
 PQS   PRS
7. SSS
8.
 PSQ   PSR
8. CPCTC
9.  PSQ
and  PSR
10.  PSQ
11.
and  PSR
PS  QR
are supp
9. Def of supp
are rt  ' s 10. If two angles are supp and
congruent, then they are
right angles.
11. Def of perpendicular
A.) (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.
A B  segm ent itself
A
B
A B  length of segm ent
Definition of Equidistance:
 If 2 points P and Q are the same distance from a third point
EQUIDISTANT
X, then X is said to be _____________
P
from P and Q.
X
Q
Definition of a
Perpendicular Bisector:
• A perpendicular bisector of a
segment is the line that both
BISECTS and is _____________
PERPENDICULAR
_________
to the segment.
L is  bisector of AB
A
B
X
L
Theorem: If 2 points are
equidistant from the endpoints of a
segment, then they determine the
P
perpendicular bisector of that
segment.
E
P 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).
Q
NEEDED: 2 points equidistant or
2 pairs congruent segments
D
Theorem: If a point is on the
perpendicular bisector of a
segment, then it is equidistant
from the endpoints of the P
segment.
If N is on PQ, the perpendicular
bisector of ED, then N is equidistant
from E and D. (EN = ND)
E
NEEDED: Perpendicular bisector
of the segment
N
Q
D
TRUE/FALSE
PRACTICE
Ready??
C
A
AD  bi sec tor of BC
E
D
B
1. E is the midpoint of BC.
TRUE
C
A
AD  bi sec tor of BC
E
D
B
2. <AEC is a right angle
TRUE
C
A
AD  bi sec tor of BC
E
D
B
3. E is the midpoint of AD
FALSE
C
A
AD  bi sec tor of BC
E
D
B
4. AC  AB
TRUE
C
A
AD  bi sec tor of BC
E
D
B
5. CE  BE
TRUE
C
A
AD  bi sec tor of BC
E
D
B
6. CA  CD
FALSE
C
A
AD  bi sec tor of BC
E
D
B
7. AE  ED
FALSE
C
A
AD  bi sec tor of BC
E
D
B
8. CB bi sec ts AD
FALSE
M
Example #1
N
Given: MN  MP
NQ  PQ
Prove: NO  PO
Statements
1. MN  MP
2. NQ  PQ
Reasons_____________________
1. Given
2. Given.
3. MQ is perpendicular to NP
3. If two points are equidistant from the
4. NO = PO
Q
O
endpoints of a segment, then they
determine the perpendicular bisector of
that segment.
4. If a point is on the perpendicular
bisector of a segment, then it is
equidistant from the endpoints of that
segment.
P
Example Problem 2
M
Given: AM≅MH
AP≅PH
Prove: ΔAPT≅ΔHPT
P
A
T
H
Statements
Reasons
1. AM≅MH
1. Given
2. AP ≅ PH
2. Given
3. MT is the perpendicular bisector
of AH
3.If two points are equidistant from the endpoints of a
segment, then they determine the perpendicular bisector of that segment.
4. < PTH, <PTA are right angles
4. Perpendicular lines form right angles (Def of perpendicular)
5. ΔPTH, ΔPTA are right triangles
5. Def of right triangle
6. PT≅PT
6. Reflexive
7. ΔAPT≅ΔHPT
7. HL (2,5,6)
Example Problem 3
K
Given: KL is the perpendicular
bisector of YE
Prove: ΔKBY≅ΔKBE
B
Y
L
E
Statements
1. KL is the perpendicular bisector
Reasons
1. Given
of YE
2. YB ≅ BE
2. If a point is on the perpendicular bisector of a
segment, then it is equidistant from the endpoints of
that segment
3. KY≅KE
3. Same as 2
4. KB≅KB
4. Reflexive
5. ΔKBY≅ΔKBE
5. SSS (2,3,4)