A Right Angle Theorem and the
Equidistance Theorems
Advanced Geometry 4.3 and 4.4
You did it! Add to your Index Cards!
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
P
Q
S
R
QR
2. Given
3. QS  SR
3. Def of midpt
4. PS  PS
5. Draw PQ and PR
4. Reflexive
6. PQ  PR
6. All radii of a circle are 
7. PQS  PRS
7. SSS
8. PSQ  PSR
8. CPCTC
9. PSQ and PSR are supp
9. Def of supp
5. Two points det a seg
10. PSQ and PSR are rt ' s 10. If two angles are supp and
congruent, then they are
right angles.
11. PS  QR
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.
AB  segment itself
A
B
AB  length of segment
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)