9572987 Geometric Constructions[1]

8/8/2019 9572987 Geometric Constructions[1]

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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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2/12

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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3/12

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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4/12

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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5/12

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

5

F

ED

C

B

A


6/12

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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7/12

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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8/12

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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9/12

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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10/12

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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11/12

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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12/12

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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9572987 Geometric Constructions[1]