Section 10.1 - Congruence Through Constructionskilmer/36609bn10.pdf · Constructing a Triangle...

c©Kendra Kilmer June 7, 2009

Section 10.1 - Congruence Through Constructions

Definitions:

• Similar (∼) objects have the same shape but not necessarily the same size.

• Congruent (∼=) objects have the same size as well as the same shape.

• A circle is the set of all points in a plane equidistant from a given point, itscenter. The distance is theradius. The word “radius” is used to describe both the segment from the center to a point on the circleand the length of that segment.

• An arc of a circle is any part of the circle that can be drawn without lifting a pencil. Thecenter of anarc is the center of the circle containing the arc.

• Given any two points on a circle, the short arc is known as theminor arc and the longer arc is knownas themajor arc . If the major arc and the minor arc are the same size, each is asemicircle. If only twopoints are used in naming an arc, the minor arc is implied.

Circle Construction:

Given a center,O, and a radius,PQ

1. Set the legs of the compass onP andQ to “measure”PQ.

2. Keeping the distance determined, set the compass pointerat the center,O, and move the pencil to drawthe circle.

Example 1: Construct a circle with center,O, and radius,PQ

s

P Qs s

O

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Segment Construction:

Given a segment,PQ

1. Set the legs of the compass onP andQ to ”measure”PQ.

2. Keeping the distance determined, place the point of the compass at any point C on linel and strike anarc to locate pointD. CD∼= PQ

Example 2: Construct a line segment congruent toPQ.

s

P Qs -� l

Triangle Congruence:△ABC is congruent to△DEF, written△ABC ∼=△DEF, if and only if 6 A ∼= 6 D,6 B∼= 6 E, 6 C ∼= 6 F, AB∼= DE, BC ∼= EF , andAC ∼= DF.

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CorrespondingParts ofCongruentTriangles areCongruent (CPCTC)

Side, Side Side Property (SSS):If three sides of one triangle are congruent, respectively,to three sides of asecond triangle, then the triangles are congruent.

Constructing a Triangle Given Three Sides:

To construct a△A′B′C′ ∼=△ABC using the three sides:

• Draw a ray−−→A′X

• Construct a segmentA′C′ ∼= AC.

• Construct a circle (or arc) with center atA′ and radiusAB and a circle (or arc) with center atC′ andradiusBC.

• Choose a point of intersection of the two circles (arcs) and label itB′

• DrawA′B′ andB′C′.

• By construction and SSS,△ABC ∼=△A′B′C′.

Example 3: Construct△A′B′C′ ∼=△ABC.

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Triangle Inequality: The sum of the measures of any two sides of a triangle must be greater than the measureof the third side.

Constructing Congruent Angles:

To copy 6 B so that one of its sides is−−→B′X :

• With center atB, mark off an arcAC to form isosceles triangleABC.

• Make an arc with same radius with center atB′.

• Label the intersection of−−→B′X and arcAC asC′.

• With pointer atC′, mark an arcC′A′ so thatC′A′ = CA.

• DrawB′A′.

• 6 B′ ∼= 6 B because△ABC ∼=△A′B′C′.

Example 4: Copy 6 B so that one of its sides is−−→B′X .

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B B′s -s

X

Question: How many different triangles can be constructed from two given segments?

Side, Angle, Side Property (SAS):

If two sides and the included angle of one triangle are congruent to two sides and the included angle ofanother triangle, respectively, then the two triangles arecongruent.

Constructions Involving Two Sides and an Included Angle of aTriangle:

To construct△A′B′C′ givenAB, AC and 6 A

• Draw a ray−−→A′X .

• CopyAC onto ray−−→A′X to getA′C′.

• Copy 6 A at A′ onA′C′ to get ray−→A′Y .

• CopyAB onto ray−→A′Y to getA′B′.

• DrawB′C′.

Example 5: Construct△A′B′C′ ∼=△ABC givenAB, AC, and 6 A:

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A B

s s

A C

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A

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Example 6: If, in two triangles, two sides and an angle not included between these sides are congruent,respectively, determine whether the triangles must be congruent.

Definitions:

• Theperpendicular bisector of a segment is a segment, ray, or line that is perpendicular to the segmentat its midpoint.

• An altitude of a triangle is the perpendicular segment from a vertex of the triangle to the line containingthe opposite side of the triangle.

Example 7: Let’s explore an isosceles triangle.

Theorem 10-1: The following holds for every isosceles triangle:

• The angles opposite the congruent sides are congruent. (Base angles of an isosceles triangle are congru-ent.)

• The angle bisector af an angle fromed by two congruent sides contains an altitude of the triangle and isthe perpendicular bisector of the third side of the triangle.

Theorem 10-2:

• Any point equidistant from the endpoints of a segment is on the perpendicular bisector of the segment.

• Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.

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Construction of the Perpendicular Bisector of a Segment:

To construct a perpendicular bisector ofAB:

• Put the compass point on A and the pencil point anywhere past the midpoint ofAB.

• Draw a circle (or arcs above and below) withA as its center.

• Keeping the distance determined, draw a circle (or arcs above and below) withB as its center.

• Draw a segment, ray, or line through the two intersection points.

Example 8: Construct the perpendicular bisector of the segmentAB

s

A Bs

Construction of a Circle Circumscribed About a Triangle:

A circle is circumscribed about a triangle when all three vertices of the triangle are on the circle.The circle is called acircumcircle. Its center is called thecircumcenter and its radius is calledthecircumradius

To construct a circle circumscribed about△ABC:

• Construct the perpendicular bisectors ofAB andAC.

• Construct a circle with radiusDA and centerD whereD is the intersection of the perpendicular bisectors.

Example 9: Construct a circle circumscribed about△ABC.

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Section 10.1 Homework Problems: 1, 3-5, 7, 9, 17-21, 24, 27-29, 35, 37, 43

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Section 10.2 - Other Congruence Properties

Angle, Side, Angle Property (ASA): If two angle and the included side of one triangle are congruent to twoangles and the included side of another triangle, respectivley, then the triangles are congruent.

Angle, Angle, Side Property (AAS): If two angles and a corresponding side of one triangle are congruentto two angles and a corresponding side of another triangle, respectively, then the two triangles arecongruent.

Example 1: Construct a triangle given the two angles and included side below:

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A B

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B

Example 2: Prove that the opposite sides of a parallelogram are congruent.

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Example 3: Prove that the opposite angles of a parallelogram are congruent.

Example 4: Prove that the diagonals of a parallelogram bisect each other.

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Example 5: Prove that the consecutive angles between parallel sides ofa trapezoid are supplementary.

Section 10.2 Homework Problems: 2, 3, 5, 6, 9, 12, 16, 26, 31, 32, 39-42

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Section 10.3 - Other Constructions

We use the definition of arhombus and the following properties (from the table in Section 10.2)to accomplish basic compass-and-straightedge constructions:

1. A rhombus is a parallelogram in which all the sides are congruent.

2. A quadrilateral in which all the sides are congruent is a rhombus.

3. Each diagonal of a rhombus bisects the opposite angles.

4. The diagonals of a rhombus are perpendicular.

5. The diagonals of a rhombus bisect each other.

Constructing Parallel Lines

Given a linel and a pointP not onl, construct a line throughP parallel tol.

Rhombus Method:

1. ThroughP, draw any line that intersectsl. Label this intersectionA. PA will be a side of therhombus.

2. Draw an arc with the pointer atA and with radiusAP to mark the third vertex,X , of the desiredrhombus.

3. With the same opening of the compass, draw intersecting arcs, first with the pointer atP and thenwith the pointer atX to findY , the fourth vertex of the rhombus.

4. Draw←→PY . We see thatPY is parallel tol as requested.

Example 1: Construct a line throughP parallel tol using the Rhombus Method.

s

P

-� l

Corresponding Angle Method:

1. ThroughP, draw a line that intersectsl.

2. Label the angle formedα.

3. Copy angleα at pointP. We see thatPQ is parallel tol as requested.

Example 2: Construct a line throughP parallel tol using the Corresponding Angle Method.

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P

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Constructing Angle Bisectors

An angle bisectoris a ray that separates an angle into two congruent angles. Toconstruct the anglebisector of6 A:

1. With the pointer atA, draw any arc intersecting the angle. Label the intersection pointsB andC. Wenow have three vertices of the rhombus:A, B, andC.

2. Draw an arc with center atB and radiusAB.

3. Draw an arc with center atC and radiusAB.

4. Label the intersection of the two arcsD, the fourth vertex of the rhombus.

5. Draw−→AD, the angle bisector of6 A.

Example 3: Construct the angle bisector of6 A.

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Example 4: Construct the angle bisector of6 A.

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Constructing Perpendicular Lines

Constructing a perpendicular to a line from a point not on the line:

1. Draw an arc with center atP that intersects the line at two points. Label these pointsA andB. Thesepoints,A, B, andP, are three of the vertices of our rhombus.

2. With the same compass opening, make two intersecting arcs, one with center atA and the otherwith center atB. Label the intersection of these arcsQ, the final vertex of our rhombus.

3. Draw←→PQ, our requested line perpendicular tol.

Example 5: Construct a line perpendicular tol throughP.

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P

-� l

Constructing a perpendicular to a line from a point on the line

1. Draw any arc with center atM that intersectsl in two points. Label the points of intersectionA andB. AB will be the diagonal of a rhombus.

2. Use a larger opening for the compass and draw intersectingarcs, with centers atA andB. Label thepoints of intersectionC andD. This is the other diagonal of our rhombus.

3. Draw←→CD, the requested line.

Example 6: Construct a line perpendicular tol throughM.

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Example 7: Construct an altitude from vertexA in the triangle below.

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Properties of Angle Bisectors

Theorem 10-3:

a) Any pointP on an angle bisector is equidistant from the sides of the angle.

b) Any point that is equidistant from the sides of an angle is on the angle bisector of the angle.

Note: The distance from a point to a line is the length of the perpendicular segment from the pointto the line.

Example 8: Prove Theorem 10-3(a).

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Constructing a Circle Inscribed in a Triangle:

Note: A line is tangent to a circle if it intersects the circle in one and only one point and isperpendicular to a radius. A circle isinscribed in a triangle if all the sides of the triangle aretangent to the circle. The inscribed circle is theincircle; the center is theincenter.

• Bisect two angles of the triangle. The intersection of the angle bisectors,P, will be the center of thecircle.

• Construct a perpendicular fromP to a side of the triangle. The length of that segment will be the lengthof the radius of the circle.

Example 9: Inscribe a circle in△ABC.

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Section 10.3 Homework Problems: 3-5, 8, 10, 21, 23-25, 41, 42

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Section 10.4 - Similar Triangles and Similar Figures

Two figures that have the same shape but not necessarily the same size aresimilar (∼).

Example 1: Use your protractor to measure each side and angle in the triangles below and summarize yourfindings.

A

B

C D

E

F

Definition of Similar Triangles: △ABC is similar to△DEF, written△ABC ∼△DEF, if and only if 6 A∼=

6 D, 6 B∼= 6 E, 6 C ∼= 6 F, andABDE

=ACDF

=BCEF

.

Note: We say that the corresponding sides areproportional and refer to the ratio of the corre-sponding side lengths as thescale factor.

Tests for Similar Triangles: We can conclude that△ABC ∼△PQR if at least one of the following condi-tions is true:

1. 6 A∼= 6 P and 6 B∼= 6 Q (AA test)

2.PQAB

=QRBC

=RPCA

(Three pairs of corresponding sides are proportional)

3.PQAB

=RPCA

and 6 A∼= 6 P (Two pairs of sides are proportional and included angles arecongruent)

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Example 2: Which of the following pairs of triangles are similar?

Example 3: In the figure below, givenBD = 12, solve forx.

Properties of Proportion:

Theorem 10-4: If a line parallel to one side of a triangle intersects the other sides, then it divides those sidesinto proportional segments.

Example 4: Prove Theorem 10-4.

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Theorem 10-5: If a line divides two sides of a triangle into proportional segments, then the line is parallel tothe third side.


Theorem 10-6: If parallel lines cut off congruent segments on one transversal, then they cut off congruentsegments on any transversal.


Construction Separating a Segment into Congruent Parts

To separateAB into a given number of congruent parts:

1. Draw any ray,−→AC, such thatA, B, andC are noncollinear.

2. Mark off the given number of congruent segments (of any size) on−→AC. In this case, we use three

congruent segments. Label the pointsA1, A2, andA3.

3. ConnectB to A3.

4. ThroughA2 andA1, construct parallels toBA3.

Example 7: SeparateAB into three congruent segments.

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Midsegments of Triangles and Quadrilaterals:

A midsegmentis a segment that connects midpoints of two adjacent sides ofa triangle or quadrilateral.

Theorem 10-7: The Midsegment Theorem:The midsegment (segment connecting the midpoints of twosides of a triangle) is parallel to the third side of the triangle and half as long.


Theorem 10-8: If a line bisects one side of a triangle and is parallel to a second side, then it bisects the thirdside and therefore is a midsegment.


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Indirect Measurements Similar triangles have long been used to make indirect measurements by usingratios involving shadows.

Example 10: On a sunny day, a tall tree casts a 40 meter shadow. At the same time, a meter stick heldvertically casts a 2.5 meter shadow. How tall is the tree?

Section 10.4 Homework Problems: 1, 2, 4, 5, 8-10, 12, 24, 25, 36, 42

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Section 10.5 - Trigonometry Ratios via Similarity

We come to the conclusion that if the angle changes, the ratiochanges too, or in other words, thisratio is the function of the angle of elevation. Since any tworight triangles with the same acuteangle are similar, the ratio of two sides of any of these similar triangles does not depend on the sizeof the triangle, it only depends on the angle.

Trigonometric Ratios

In any right triangleABC with a right angle atC, we refer tob = AC as the side adjacent to6 A,a = BC as the side opposite6 A, andc = AB as the hypotenuse. Each of the ratios

side opposite6 Ahypotenuse

side adjacent6 Ahypotenuse

side opposite6 Aside adjacent6 A

does not depend on the size of the triangle, it depends only onthe size of6 A. In a right triangleABC with 6 C = 90◦, we define

• sin( 6 A) =side opposite6 A

hypotenuse

• cos( 6 A) =side adjacent6 A

hypotenuse

• tan( 6 A) =side opposite6 Aside adjacent6 A

Example 1: For each of the following figures, solve forx

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Note: You are responsible for remembering the sine, cosine,and tangent of 3 special angles, 30◦,45◦, and 60◦.

Example 2: Let’s look at the sin, cos, and tan of these 4 special angles.

Example 3: For each of the following figures, find the exact value ofx.

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Example 4: Show that tan( 6 A) =sin( 6 A)

cos( 6 A)

Example 5: Use the Pythagorean Theorema2+b2 = c2 (herea andb are the legs andc is the hypotenuse ofa right triangle) to show that sin2 6 A+cos2 6 A = 1.

Section 10.5 Homework Problems: 1-4, 9, 14

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Section 10.6 - Lines in a Cartesian Coordinate System

A Cartesian Coordinate Systemis constructed by placing two number lines perpendicular toeachother. The intersection point of the two lines is theorigin , the horizontal line is thex-axis, and thevertical line is they-axis. The location of any point can be described by an ordered pairof numbers(a,b). The first component is thex-coordinate; the second component is they-coordinate.

Example 1: Plot the following points on a Cartesian Coordiante System.

Equations of Vertical and Horizontal Lines

• Vertical Line: The graph of the equationx = a, wherea is some real number, is a line perpendicular tothex-axis through the point with coordinates(a,0).

• Horizontal Line: The graph of the equationy = b, whereb is some real number, is a line perpendicularto they-axis through the point with coordinates(0,b).

Example 2: Sketch the graph for each of the following equations:

Questions: What is the equation of thex-axis? What is the equation of they-axis?

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Equations of Lines

Every line has an equation of the form eithery = mx+b or x = a, wherem is the slope andb is they-intercept.

Example 3: Find the equation of the line with slope−2 andy-intercept 4.

Slope Formula: Given two pointsA(x1,y1) andB(x2,y2) with x1 6= x2, the slopem of the line←→AB is

m =y2− y1

x2− x1

Note: Any two parallel lines have the same slope, or are vertical lines with undefined slope.

Example 4: Find the slope of the line passing through the points(−4,6) and(2,−3). Graph the line.

Point-Slope Form of the Equation of the Line: The equation of the line that passes through the point(x1,y1) and has slopem is given by:

y− y1 = m(x− x1)

Example 5: Find the equation of the line passing through the points(−4,6) and(2,−3).

Example 6: Find thex andy intercepts of the line with equationy =−3x+8.

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Systems of Linear Equations

What are the possible cases when solving a system of linear equations?

Example 7: Solve the following system of linear equations:

4x−5y =−30

2x+ y =−8


2x− y = 1

6x−3y = 12

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3x−7y = 4

6x−14y = 8

Example 10: Diego’s piggy bank contains 55 coins. If all of the coins are either nickels or dimes and thevalue of the coins is $4.75, how many of each kind of coin are there?

Section 10.6 Homework Problems: 1-7, 9, 10, 22, 24, 26, 43

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