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Construction Reference
Construct the perpendicular bisector of a line segment.
Or, construct the midpoint of a line segment.
1. Begin with line segmentXY. YX
2. Place the compass at pointX. Adjust the compass radiusso that it is more than ()XY. Draw two arcs as shown
here.YX
3. Without changing the compass radius, place the compass
on point Y. Draw two arcs intersecting the previously
drawn arcs. Label the intersection pointsA andB.
A
B
YX
4. Using the straightedge, draw lineAB. Label the
intersection point M. Point Mis the midpoint of line
segmentXY, and lineAB is perpendicular to linesegmentXY.
X Y
A
B
M
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Given point P on line k, construct a line through P, perpendicular to k.
1. Begin with line k, containing pointP. k
P
2. Place the compass on pointP. Using an arbitrary radius,
draw arcs intersecting line kat two points. Label the
intersection pointsXand Y.
k
P YX
3. Place the compass at pointX. Adjust the compass radius
so that it is more than ()XY. Draw an arc as shown
here.k
P YX
4. Without changing the compass radius, place the compass
on point Y. Draw an arc intersecting the previously
drawn arc. Label the intersection pointA. k
A
P YX
5. Use the straightedge to draw lineAP. LineAPisperpendicular to line k.
k
A
X YP
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Given point R, not on line k, construct a line through R, perpendicular to k.
1. Begin with point line kand pointR, not on the line.k
R
2. Place the compass on pointR. Using an arbitrary radius,
draw arcs intersecting line kat two points. Label theintersection pointsXand Y.
kR
YX
3. Place the compass at pointX. Adjust the compass radius
so that it is more than ()XY. Draw an arc as shown
here.
k
YX
R
4. Without changing the compass radius, place the compasson point Y. Draw an arc intersecting the previously
drawn arc. Label the intersection pointB.
k
B
YX
R
5. Use the straightedge to draw lineRB. LineRB is
perpendicular to line k.
k
B
YX
R
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Construct the bisector of an angle.
1. Let pointPbe the vertex of the angle. Place the
compass on pointPand draw an arc across both sidesof the angle. Label the intersection pointsQ andR.
Q
RP
2. Place the compass on point Q and draw an arc across
the interior of the angle. Q
RP
3. Without changing the radius of the compass, place iton pointR and draw an arc intersecting the one drawn
in the previous step. Label the intersection pointW.
P
Q
R
W
4. Using the straightedge, draw rayPW. This is the
bisector ofQPR.
P
Q
R
W
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Construct an angle congruent to a given angle.
1. To draw an angle
congruent to A, begin
by drawing a ray with
endpointD.
D
A
2. Place the compass on
pointA and draw an arcacross both sides of the
angle. Without
changing the compassradius, place the
compass on pointD anddraw a long arccrossing the ray. Label
the three intersection
points as shown.
E
B
C
D
A
3. Set the compass so thatits radius isBC. Place
the compass on pointE
and draw an arcintersecting the one
drawn in the previousstep. Label the
intersection pointF.
F
ED
C
B
A
4. Use the straightedge todraw rayDF.
EDF BACF
ED
C
B
A
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F
ED
C
B
A
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Given a line and a point, construct a line through the point, parallel to the given line.
1. Begin with pointPand line k.
k
P
2. Draw an arbitrary line through pointP,
intersecting line k. Call the intersection pointQ.
Now the task is to construct an angle with vertexP, congruent to the angle of intersection.
kQ
P
3. Center the compass at pointQ and draw an arcintersecting both lines. Without changing the
radius of the compass, center it at pointPand
draw another arc.
kQ
P
4. Set the compass radius to the distance between
the two intersection points of the first arc. Nowcenter the compass at the point where the second
arc intersects linePQ. Mark the arc intersection
pointR. k
R
Q
P
5. LinePR is parallel to line k.
k
R
Q
P
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Given a line segment as one side, construct an equilateral triangle.
This method may also be used to construct a 60 angle.
1. Begin with line segment TU.
UT
2. Center the compass at point T, and set thecompass radius to TU. Draw an arc as shown.
UT
3. Keeping the same radius, center the compass atpoint Uand draw another arc intersecting the
first one. Let point Vbe the point of
intersection.
V
UT
4. Draw line segments TVand UV. Triangle TUV
is an equilateral triangle, and each of its interior
angles has a measure of 60.
V
UT
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Divide a line segment into n congruent line segments.
In this example, n = 5.
1. Begin with line segmentAB. It will bedivided into five congruent line segments.
BA
2. Draw a ray from pointA. Use the compass
to step off five uniformly spaced points along
the ray. Label the last point C.
C
BA
3. Draw an arc with the compass centered at
pointA, with radiusBC. Draw a second arc
with the compass centered at pointB, withradiusAC. Label the intersection pointD.
Note thatACBD is a parallelogram.
D
C
BA
4. Use the compass to step off points along line
segmentDB, using the same radius that wasused for the points along line segmentAC.
D
C
BA
5. Use the straightedge to connect thecorresponding points. These line segments
will be parallel. They cut line segmentsAC
andDB into congruent segments. Therefore,they must also cut line segmentAB intocongruent segments.
D
C
BA
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Given a circle, its center point, and a point on the exterior of the circle, construct a line
through the exterior point, tangent to the circle.
1. Begin with a circle centered on point C. PointPis on the exterior of the circle. PC
2. Draw line segment CP, and construct point M,
the midpoint of line segment CP. (For theconstruction of the midpoint, refer to the
perpendicular bisector construction, on page 1.)C
P
M
3. Center the compass on point M. Draw a circlethrough points CandP. It will intersect the
other circle at two points,R and S.
R
S
P
MC
4. PointsR and Sare the tangent points. LinesPR
andPSare tangent to the circle centered onpoint C.
S
R
P
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Construct the center point of a given circle.
1. Begin with a circle, but no center point.
2. Draw chordAB. A
B
3. Construct the perpendicular bisector of
chordAB. Let CandD be the pointswhere it intersects the circle. (Refer to
the construction of a perpendicular
bisector, on page 1.) C
D
A
B
4. Chord CD is a diameter of the circle.
Construct pointP, the midpoint ofdiameterCD. PointPis the center pointof the circle. (Refer to the construction
of the midpoint of a line segment, on
page 1.)
PD
C
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Given three noncollinear points, construct the circle that includes all three points.
1. Begin with pointsA,B, and C. C
B
A
2. Draw line segmentsAB andBC. C
B
A
3. Construct the perpendicular bisectors of line
segmentsAB andBC. (Refer to the
perpendicular bisector construction, on page 1.)
Let pointPbe the intersection of theperpendicular bisectors.
A
B
CP
4. Center the compass on pointP, and draw the
circle through pointsA,B, and C.A
B
CP
Given a triangle, circumscribe a circle.
1. Begin with triangle STU.
U
T
S
2. If a circle is circumscribed around the triangle,then all three vertices will be points on the circle,
so follow the instructions above, for construction
of a circle through three given points.
S
T
U
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Given a triangle, inscribe a circle.
1. Begin with triangleKLM.
M
L
K
2. Construct the bisectors ofKand L.
(Refer to the angle bisector construction, onpage 4.) Let pointQ be the intersection of
the two angle bisectors.Q
M
L
K
3. Construct a line through point Q,
perpendicular to line segmentKL. Let pointR be the point of intersection. (Refer to the
construction of a perpendicular line through
a given point, on page 3.)
R
Q
M
L
K
4. Center the compass on pointQ, and draw a
circle through pointR. The circle will betangent to all three sides of a triangle. R
Q
M
L
K
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