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Chapter 10 Constructions

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**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

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**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

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**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.

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**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

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**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.

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**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.

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**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.

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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”

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**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

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**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

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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

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**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

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**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

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**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!!

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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

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**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

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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.

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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,…)

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**Examples of a Locus in a Plane**

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Locus Problems

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