Section 10.1 – Congruence Through Constructions
Definitions
• Similar (∼) objects have the same shape but not necessarily the same size.
• Congruent (≅) objects have the same shape and same size.
• A circle is the set of all points in a plane equidistant from a given point, its center. The
distance is the radius. The word “radius” is used to describe both the segment from the
center to a point on a circle and the length of that segment.
• An arc of a circle is any part of the circle that can be drawn without lifting a pencil. The
center of an arc is the center of the circle containing the arc.
• Given any two points on a circle, the short arc is known as the minor arc and the longer arc
is known as the major arc. If the major arc and the minor arc are the same size, each is a
semicircle. If only two points 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 on P and Q to “measure” PQ .
2. Keeping the distance determined, set the compass pointer at the center, O, and move the
pencil to draw the circle.
Example: Construct a circle with center, O, and radius, PQ .
P
Q
O
Segment Construction
Given a segment PQ
1. Set the legs of the compass on P and Q to “measure” PQ .
2. Keeping the distance determined, place the point of the compass at any point C on line l and
strike an arc to locate point D. CD ! PQ
Example: Construct a line segment on l congruent to PQ .
P
l
Q
1
Triangle Congruence
Triangle ABC is congruent to triangle DEF, written ABC ≅ DEF, if and only if ∠A ≅ ∠D,
∠B ≅ ∠E, ∠C ≅ ∠F, AB ! DE , BC ! EF , and AC ! DF . In other words, Corresponding
Parts of Congruent Triangles are Congruent (CPCTC).
Side, Side, Side Property (SSS)
If the three sides of one triangle are congruent, respectively, to the three sides of a second
triangle, then the triangles are congruent.
Constructing a Triangle Given Three Sides
To construct A’B’C’ ≅ ABC using the three sides,
1. Construct a segment A'C ' ! AC .
2. Construct a circle (or arc) with center at A’ and radius AB, and a circle (or arc) with
center at C’ and radius BC.
3. Label one of the intersection points of the two circles (or the intersection of the arcs) B’.
4. Draw A' B' and B'C ' .
Example: Construct A’B’C’ ≅ ABC.
B
A
C
2
Triangle Inequality Property
The sum of the measures of any two sides of a triangle must be greater than the measure of the
third side.
Constructing Congruent Angles
!!!"
To copy ∠B so that one of its sides is B!X :
1. With center at B, mark off an arc AC to form isosceles triangle ABC.
2. Make an arc with same radius with center at B′.
!!!"
3. Label the intersection of B!X and arc AC as C′.
4. “ Measure” CA. With pointer at C′, mark an arc C′A′ so that C′A′ = CA.
5. Draw B′A′.
6. ∠B′ ≅ ∠B because △ABC ≅ △ A!B!C ! .
!!!"
Example: Copy ∠B so that one of its sides is B!X .
B
B′
X
Side, Angle, Side Property (SAS)
If two sides and the included angle of one triangle are congruent to two sides and the included
angle of another triangle, respectively, then the two triangles are congruent.
Constructions Involving Two Sides and an Included Angle of a Triangle
To construct △ A!B!C ! given AB , AC and ∠A.
!!!"
1. Draw a ray A!X .
!!!"
2. Copy AC onto ray A!X to get A!C ! .
!!!"
3. Copy ∠A at A′ on A!C ! to get ray A!Y .
!!!"
4. Copy AB onto ray A!Y to get A!B! .
5. Draw B!C ! .
3
Example: Construct △ A!B!C ! given AB , AC and ∠A.
A
A
B
C
Definitions
• The perpendicular bisector of a segment is a segment, ray, or line that is perpendicular to
the segment at its midpoint.
• An altitude of a triangle is the perpendicular segment from a vertex of the triangle to the
line containing the opposite side of the triangle.
Example: Consider the altitude and perpendicular bisector of an isosceles triangle.
Theorem 10-1
For every isosceles triangle:
a. The angles opposite to the congruent sides are congruent. (Base angles of an isosceles
triangle are congruent.)
b. The angle bisector of an angle formed by two congruent sides contains an altitude of the
triangle and is the perpendicular bisector of the third side of the triangle.
Theorem 10-2
a. Any point equidistant from the endpoints of a segment is on the perpendicular bisector of
the segment.
b. 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 of AB
1. Put the compass point on A and the pencil point anywhere past the midpoint of AB .
2. Draw a circle (or arcs above and below) with A as its center.
3. Keeping the distance determined, draw a circle (or arcs above and below) with B as its
center.
4. Draw a segment, ray, or line through the two intersection points.
Example: Construct the perpendicular bisector of the segment AB .
A
B
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 a circumcircle. Its center is called the circumcenter and its radius the
circumradius.
To construct a circle circumscribed about △ABC:
1. Construct the perpendicular bisectors of AB and AC .
2. Construct a circle with radius DA and center D where D is the intersection of the
perpendicular bisectors.
Example: Construct a circle circumscribed about △ABC.
B
C
A
5
Section 10.2 – Other Congruence Properties
Angle, Side, Angle Property (ASA)
If two angles and the included side of one triangle are congruent to two angles and the included
side of another triangle, respectively, then the triangles are congruent.
Angle, Angle, Side Property (AAS)
If two angles and a corresponding side of one triangle are congruent to two angles and a
corresponding side of another triangle, respectively, then the two triangles are congruent.
Example: Construct a triangle given the two angles and included side below.
A
B
A
B
Example: Prove the opposite sides of a parallelogram are congruent.
D
A
C
B
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Example: Prove the diagonals of a parallelogram bisect each other.
D
A
C
B
Example: Prove the consecutive angles between parallel sides of a trapezoid are supplementary.
A
D
B
C
7
Section 10.3 – Other Constructions
From 10.2, a rhombus is a parallelogram in which all the sides are congruent. We’ll use the
rhombus as the basis of more constructions.
Properties of a Rhombus
1. A quadrilateral in which all the sides are congruent is a rhombus.
2. Each diagonal of a rhombus bisects the opposite angles.
3. The diagonals of a rhombus are perpendicular.
4. The diagonals of a rhombus bisect each other.
Constructing Parallel Lines
Given a line l and a point P not on l, construct a line through P parallel to l.
Rhombus Method
1. Through P, draw any line that intersects l. Label the intersection A. PA will be a side of a
rhombus.
2. Draw an arc with the pointer at A and radius AP to mark the third vertex, X, of a
rhombus.
3. With the same opening of the compass, draw intersecting arcs, first with the pointer at P
and then with the pointer at X to find Y, the fourth vertex of the rhombus.
!##" !##"
4. Draw PY . PY || l in rhombus APYX.
Example: Construct a line through P parallel to l by using the rhombus method.
P
l
Corresponding Angle Method
1. Through P, draw a line that intersects l, forming angle α.
!##"
2. Copy angle α at point P. PY // l. (Recall that copying an angle was covered in 10.1)
Example: Construct a line through P parallel to l by using the corresponding angle method.
P
l
8
Constructing Angle Bisectors
An angle bisector is a ray that separates an angle into two congruent angles.
1. Given an angle A, with the pointer at A, draw any arc intersecting the angle at B and C,
giving three vertices of a rhombus with vertex at A.
2. Draw an arc with center at B and radius AB.
3. Draw an arc with center at C and radius AB. The arcs intersect at D, the fourth vertex of
the rhombus.
!!!"
4. Connect A with D. AD is the angle bisector of A in rhombus ABDC. (The diagonals of a
rhombus bisect the angles.)
Example: Construct an angle bisector of ∠ A.
A
Constructing Perpendicular Lines
The distance from a point to a line is the length of the perpendicular segment from the point to the line.
Construction of a Perpendicular to a Line through a Point not on the Line
1. Draw an arc with center at P that intersects the line at two points, A and B. (P, A, and B
will be vertices of a rhombus.)
2. With the same compass opening, make two intersecting arcs, one with center at A and the
other with center at B. Label the intersection Q, the fourth vertex of the rhombus.
!##"
3. Connect P with Q. PQ is perpendicular to l.
Example: Construct a line through P (not on l), perpendicular to line l.
P
l
9
Construction of a Perpendicular to a Segment through a Point on the Segment
1. Draw an arc with center at M that intersects l in two points, A and B. ( AB will be the
diagonal of a rhombus.)
2. Use a larger opening for the compass and draw intersecting arcs, with centers at A and B,
to determine C and D (endpoints of the other diagonal of the rhombus).
!##"
3. Connect C with D, the points where the arcs intersect. CD is perpendicular to l through
!##"
!##"
M. ( CD and AB are perpendicular bisectors of each other in rhombus ACBD.)
Example: Construct a line through M (on l), perpendicular to line l.
Example: Construct an altitude from vertex A in the triangle below.
A
B
C
10
Properties of Angle Bisectors
Theorem 10-3
a. Any point P 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.
Example: Prove Theorem 10-3a.
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Definitions
• A line is tangent to a circle if it intersects the circle in one and only one point and is
perpendicular to a radius.
• A circle is inscribed in a triangle if all the sides of the triangle are tangent to the circle.
o The inscribed circle is the incircle; the center is the incenter.
Constructing a Circle Inscribed in a Triangle.
1. Bisect the angles of the triangles. The intersection of the angle bisectors, P, will be the
center of the circle.
2. Construct a perpendicular from P to a side of the triangle. The length of that segment will
be the length of the radius of the circle.
Example: Inscribe a circle in ABC.
A
B
C
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Section 10.4 – Similar Triangles and Similar Figures
Similar Triangles
Two figures that have the same shape but not necessarily the same size are similar (~). Corresponding
sides are proportional and the ratio of the corresponding side lengths is the scale factor.
Example: Are ABC and DEF similar?
B
E
D
A
F
C
Definition of Similar Triangles
ABC is similar to DEF (ABC ∼ DEF) iff ∠ A ≅ ∠ D, ∠ B ≅ ∠ E, ∠ C ≅ ∠ F, and
AB
AC BC
=
=
.
DE DF
EF
Tests for Similar Triangles: We can conclude that △ABC ∼ △PQR if at least one of the
following conditions is true:
1. ∠A ≅ ∠P and ∠B ≅ ∠Q (Angle-Angle or AA Property)
PQ QR RP
=
=
2.
(Three pairs of corresponding sides are proportional)
AB BC CA
PQ RP
=
3.
and ∠A ≅ ∠P (Two pairs of sides are proportional and included angles are congruent)
AB CA
Example: In the figure below, given BD = 12, solve for x.
D
x
A
8
E
C
5
B
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Properties of Proportion
Theorem 10-4
If a line parallel to one side of a triangle intersects the other sides, then it divides those sides into
proportional segments.
Example: Prove Theorem 10-4.
B
x
z
m
y
w
A
C
Theorem 10-5
If a line divides two sides of a triangle into proportional segments, then the line is parallel to the
third side.
Example: Prove Theorem 10-5.
B
x
y
A
z
m
w
C
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Theorem 10-6
If parallel lines cut off congruent segments on one transversal, then they cut off congruent
segments on any transversal.
B
a
b
c y
d
A
e
f
g
h
C
Construction Separating a Segment into Congruent Parts
!!!"
1. Draw any ray, AC , such that A, B, and C are noncollinear.
!!!"
2. Mark off the given number of congruent segments (of any size) on AC . (For the
example, we’ll use three congruent segments.)
3. Connect B to A3, the endpoint of the last of the three congruent segments.
4. Through A2 and A1, construct parallels to BA3 .
Example: Divide AB into three congruent parts.
A
B
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Midsegments of Triangles and Quadrilaterals
A midsegment is a segment that connects midpoints of two adjacent sides of a triangle or quadrilateral.
Theorem 10-7
The Midsegment Theorem The midsegment is parallel to the third side of the triangle and half as long.
Example: Prove Theorem 10-7.
B
D
E
A
C
Theorem 10-8
If a line bisects one side of a triangle and is parallel to a second side, then it bisects the third side
and therefore is a midsegment.
Example: Prove Theorem 10-8.
B
D
C
A
E
Indirect Measurements
Similar triangles and rations involving shadows can be used to make indirect measurements
(such as determining the height of the Great Pyramid of Egypt).
Example: On a sunny day, a tall tree casts a 60 meter shadow. At the same time, a meter stick held
vertically casts a 3 meter shadow. How tall is the tree?
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Section 10.5 – Trigonometry Ratios via Similarity
Any two right triangles with the same acute angle are similar, so the ratio of two sides of any of
these similar triangles does not depend on the size of the triangle; it only depends on the angle.
Trigonometry applies properties of right triangles and similar triangles to describe angle and
distance relationships.
Trigonometric Ratios
The hypotenuse of a right triangle is the longest side and is opposite the right angle.
Trigonometric functions are defined as ratios of the lengths of the sides of a right triangle. You
are responsible for the trig function values of 30°, 45°, and 60° angles.
sin(!A) =
a length of opposite side opp
=
=
c
length of hypotenuse
hyp
cos(!A) =
b length of adjacent side adj
=
=
c
length of hypotenuse
hyp
tan(!A) =
a length of opposite side opp
=
=
b length of adjacent side
adj
Example: Solve for x.
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Pythagorean Theorem: In a right triangle, with hypotenuse c and sides a and b, c2 = a2 + b2.
In a 30° - 60° - 90° triangle, the length of the side opposite the 30° angle is half the length of the
hypotenuse.
30-60-90 Triangle
1
2
1
60°
160°
2
60°
1
30°
1
60° 2
30°
30°
3
2
2
1
30°
3
2
45-45-90 Triangle
2
2
45°
45°
1
1
2 45°
2
45°
1
Example: Find the exact value of x.
3
3
2
18
Example: Find the exact value of y.
Example: Show that tan(!A) =
sin(!A)
.
cos(!A)
19
Section 10.6 – Lines in a Cartesian Coordinate System
The Cartesian coordinate system is constructed by placing two number lines perpendicular to
each other.
• The intersection point of the two lines is the origin, the horizontal lines is the x-axis; the
vertical line is the y-axis.
• An ordered pair of numbers describes the location of any point. The first component in
the ordered pair is the x-coordinate; the second component is the y-coordinate.
Example: Plot and label the points P(2, −3) and Q( −4, 0) on the Cartesian coordinate system.
Equations of Vertical and Horizontal Lines
•
•
Vertical Lines
o The graph of the equation x = a, where a is some real number, is a line
perpendicular to the x-axis through the points with coordinates (a, 0).
Horizontal Lines
o The graph of the equation y = b is a line perpendicular to the y-axis through the
point with coordinates (0, b).
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Example: Graph the following:
x = −3
y=3
What is the equation of the x-axis? The y-axis?
Equations of Lines
Every line has an equation of the form either y = mx + b or x = a, where m is the slope and (0, b)
is the y-intercept.
Example: Find the equation of the line with slope −2 and y-intercept 4.
Slope Formula
!##"
Given two points A (x1, y1) and B(x2, y2) with x1 ≠ x2, the slope m of the line AB is
m=
rise !y y2 " y1
=
=
run !x x2 " x1
Any two parallel lines have the same slope, or are vertical lines with undefined slope.
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Example: Find the slope of the line through (−4, 6) and (2, −3). Graph the line.
Point-Slope Form
The equation of the line that passes through the point (x1, y1) with slope m is given by:
y – y1 = m(x – x1)
Example: Find the equation of the line through (−4, 6) and (2, −3).
Example: What are the x- and y-intercepts of the line y = −3x + 8?
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Systems of Linear Equations
What are the three possible outcomes for the solution to a system of linear equations?
Example: Solve the system:
4x – 5y = –30
2x + y = –8
Example: Solve the system:
2x – y = 1
6x – 3y = 12
Example: Solve the system:
3x – 7y = 4
6x – 14y = 8
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