1. Chapter 10
Constructions

2. Don’t change your radius!
Construction #1
Construct a segment congruent to a given segment.
This is our compass.
Given: A B
Procedure:
1. Use a straightedge to draw a line. Call it l.
Construct: XY = AB
Don’t change your radius!
2. Choose any point on l and label it X.
3. Set your compass for radius AB and make a mark on the line where B lies. Then, move your compass to line l and set your pointer on X. Make a mark on the line and label it Y. l X Y

3. Construct an angle congruent to a given angle
Construction #2
Construct an angle congruent to a given angle A C B
Given:
Procedure: D
1) Draw a ray. Label it RY.
2) Using B as center and any radius, draw an arc intersecting BA and BC. Label the points of intersection D and E. E
Construct:
3) Using R as center and the SAME RADIUS as in Step 2, draw an arc intersecting RY. Label point E2 the point where the arc intersects RY D2 R Y
4) Measure the arc from D to E. E2
5) Move the pointer to E2 and make an arc that that intersects the blue arc to get point D2
6) Draw a ray from R through D2

4. Bisector of a given angle?
Construction #3
How do I construct a
Bisector of a given angle? C A B Z
Given: X Y
Procedure:
Using B as center and any radius, draw
and arc that intersects BA at X and BC at point Y.
2. Using X as center and a suitable radius, draw and arc.
Using Y as center and the same radius, draw an arc
that intersects the arc with center X at point Z.
3. Draw BZ.

5. How do I construct a perpendicular bisector to a given segment?
Construction #4
How do I construct a perpendicular
bisector to a given segment? X
Given: A B Y
Procedure:
Using any radius greater than 1/2 AB, draw four arcs of
equal radii, two with center A and two with center B.
Label the points of intersection X and Y.
Draw XY

6. Construction #5 How do I construct a perpendicular bisector
to a given segment at a given point? Z
Given: C k X Y
Procedure:
Using C as center and any radius, draw arcs intersecting
k at X and Y.
Using X as center and any radius greater than CX, draw an arc. Using Y as center and the same radius,
draw and arc intersecting the arc with center X at Z.
Draw CZ.

7. Construction #6 How do I construct a perpendicular bisector to a
given segment at a given point outside the line? k P
Given: X Y Z
Procedure:
Using P as center, draw two arcs of equal radii that
intersect k at points X and Y.
Using X and Y as centers and a suitable radius, draw arcs
that intersect at a point Z.
Draw PZ.

8. Construction #7 How do I construct a line parallel to a
given line through a given point? k P 1 l
Given: A B
Procedure:
Let A and B be two points on line k. Draw PA.
At P, construct <1 so that <1 and <PAB are congruent
corresponding angles. Let l be the line containing the
ray you just constructed.

9. Concurrent Lines
If the lines are Concurrent then they all intersect at the same point.
The point of intersection is called the “point of concurrency”

10. What are the 4 different types of concurrent lines for a triangle?
Perpendicular bisectors
Angle bisectors
Altitudes
Medians
Circumcenter
Incenter
Orthocenter
Centriod
Concurrent Lines
Point of Concurrency
Circumcenter SP
Orthocenter SP
Incenter SP
Centroid SP

11. Given a point on a circle, construct the tangent to the circle at the given point .
Construction #8
PROCEDURE:
Given: Point A on circle O.
1) Draw Ray OA
2) Construct a perpendicular through OA at point A. 5 3
Now, using the same radius, construct arcs 5 & 6 using point P as the center so that they intersect arcs 3 & 4 to get points X & Y
Construct arcs 3 & 4 using point Q as the center and any suitable radius (keep this radius) X
3) Draw tangent line XY 2 1 O A Q P
Construct arcs 1 and 2 using any suitable radius and A as the center Y 6 4

12. Construction #9
Given a point outside a circle, construct a tangent to the circle from the given point.
PROCEDURE:
Given: point A not on circle O
1) Draw OA. X 3 1
2) Find the midpoint M of OA (perpendicular bisector of OA)
3) Construct a 2nd circle with center M and radius MA M O A
4) So you get points of tangency at X & Y where the arcs intersect the red circle
Construct arcs 1& 2 using a suitable radius greater than ½AO
( keep this radius for the next step)
Construct arcs 3& 4 using the same radius
(greater than ½AO)
You get arcs 5 & 6
5) Draw tangents AX & AY Y 4 2

13. Construction #10 Given a triangle construct the circumscribed circle.
PROCEDURE:
Given: Triangle ABC 7 8 A B C 5 6
1) Construct the perpendicular bisectors of the sides of the triangle and label the point of intersection F. 3 4 1 2
Bisect segment BC; Using a radius greater than 1/2BC from point C construct arcs 5 & 6
From point B construct arcs 7 & 8 and draw a line connecting the intersections of the arcs
2) Set your compass pointer to point F and the radius to measure FC. F
3) Draw the circle with center F , that passes through the vertices A, B, & C
radius
Now construct the perpendicular bisector of segment AB and label point F, where the 3 lines meet.
Bisect segment AC; Using a radius greater than 1/2AC from point C construct arcs 1 & 2
From point A construct arcs 3 & 4 and draw a line connecting the intersections of the arcs

14. Construction #11 Given a triangle construct the inscribed circle.
PROCEDURE:
Given: Triangle ABC A B C
1) Construct the angle bisectors of angles A, B, & C, to get a point of intersection and call it F
2) Construct a perpendicular to side AC from point F, and label this point G. F
3) Put your pointer on point F and set your radius to FG.
4) Draw the circle using F as the center and it should be tangent to all the sides of the triangle. G X Y

15. Three congruent segments
Remember you made 3 because you are dividing by 3, but if you wanted to divide by, say, 6 you would have to make 6 congruent parts on the ray and so on for 7,8,9…
Construction #12
Given a segment, divide the segment into a given number of congruent parts.
Given: Segment AB
PROCEDURE:
Divide AB into 3 congruent parts.
1) Construct ANY RAY from point A that’s not AB A B D C
2) Construct 3 congruent segments on the ray using ANY RADIUS starting from point A. Label the new points X, Y, & Z X
3) Draw segment ZB and copy the angle AZB ( 1) to vertices X & Y
So AC=CD=DB
Three congruent segments Y 1
Use any suitable radius that will give some distance between the points
4) Draw the the rays from X & Y, they should be parallel to the segment ZB and divide AB into 3 congruent parts. Z
Remember keep the same radius!!

16. Construction #13
Given three segments construct a fourth segment (x) so that the four segments are in proportion. a b c
Given:
Construct: segment x such that a c x b b x
PROCEDURE: a
1) Using your straight edge construct an acute angle of any measure. 1 N
2) On the lower ray construct “a” and then “c “ from the end of “a”. c
3) On the upper ray construct “b” and then connect the ends of “a & b”
4) Next copy angle 1 at the end of “c” and then construct the parallel line

17. Construction #14 X Given 2 segments construct their geometric mean.
Make sure to set your radius to more
than 1/2NM then:
Mark off 2 arcs from M. Keep the same
radius and mark off 2 more arcs from N,
crossing the first two.
Draw the perpendicular bisector PQ
through point O
Mark off 2 arcs from Y.
Keep the same radius and mark off 2 more arcs from X, crossing the first two. a b
Given:
Construct: segment x such that:
Mark off 2 equal distances on either
side of point O using any radius and
then bisect this new segment
(I used the distance from O to M,
but remember any radius will do) a x b
The orange segment is x the geometric mean between the lengths of a and b
PROCEDURE: 9 7 3 1 X Q K
1) Draw a ray and mark off a+b.
2) Bisect a+b (XY) and label point M. 6 5 X Y M
3) Construct the circle with center M and radius = MY (or MX) a N O b
4) Construct a perpendicular where segment “a” meets segment “b” (point O) 8 10 2 4 P L

18. The Meaning of Locus
If a figure is a locus then it is the set of all points that satisfy one or more conditions.
The term “locus” is just a technical term meaning “a set of points”.
So , a circle is a locus.
Why??
Because it is a set of points a given distance from a given point.

19. What can a locus be??
Remember it is a SET OF POINTS so if you recall the idea of sets from algebra it is possible for a set to be empty.
So a set could be:
The empty set. (no points fit the condition or conditions)
A single point.
Two points, three points….
An infinite set of points. (like a line, circle, curve,…)

20. Examples of a Locus in a Plane

21. Locus Problems