Preparing for the Regents Examination Geometry, AK

5 1. −−
FG is the perpendicular bisector of −−
HI .
2. −−
JG is the perpendicular bisector of −−
HI .
(F, J, and G are collinear.)
3. −−
HJ −
IJ
4. −−
GJ −−
GJ
5. HJG and IJG are right angles.
(Definition of perpendicular bisector)
6. HJG IJG (Right angles are
congruent.)
7. HGJ IGJ (SAS SAS)
8. 1 2 (Corresponding parts of
congruent triangles are congruent.)
6 1. −−
AB and −−−
CD bisect each other.
2. −−
AB −−−
CD
3. −−
AC −−−
AD
4. −−
BC −−
BD
5. −−−
AD −−
BD
6. −−
AC −−
BC
7. ACBD is an equilateral quadrilateral.
(Definition of equilateral quadrilateral)
7 1. 1 2
2. 3 4
3. −−
XZ −−
XZ
4. XVZ XWZ (ASA ASA)
5. −−−
XV −−−
XW (Corresponding parts
of congruent triangles
are congruent.)
6. XV XW
7. −−−
XZY ; −−−−
VYW
8. −−
XY is the perpendicular bisector of −−−−
VYW .
(A point equidistant from the endpoints
of a line segment is on the perpendicular
bisector of the line segment.)
8 BP CP
4 y 4
y 1
AP CP
x y 4
x 1 4
x 5
9 AP BP
3x y x y
x y
BP CP
x y 4
x x 4
x 2
y 2
10 CG AG 10
24 Chapter 5: Congruence Based on Triangles
5-7 Constructions
(page 83)
1 (4) BAC and GHI
2 A B C
3
4
5
S
A B
V
E F
T
C D
R
1 2
3
6 If the legs of the compass form two sides of
a triangle, then the line segment connecting
the pencil to the point is the third side. If
the angle between the legs does not change
(and the legs remain the same length), then
any line segment connecting the legs is
congruent to the original because of SAS
congruence.
7
A
G
F
B
E
H C
D L
The first arc drawn in the construction creates
an isosceles triangle, BGH. The rest of
the construction copies the sides of BGH to
a new triangle. ED BH, FD GH, and
FE GB, since F and G lie on the intersections.
Therefore, BGH EFD by SSS
congruence and all corresponding angle
measures, including B and E, are equal.