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Lesson 5-1. Bisectors, Medians and Altitudes. Transparency 5-1. 5-Minute Check on Chapter 4. Refer to the figure. 1. Classify the triangle as scalene, isosceles, or equilateral. 2. Find x if m A = 10 x + 15, m B = 8 x – 18, and m C = 12 x + 3. - PowerPoint PPT Presentation
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Lesson 5-1
Bisectors, Medians and Altitudes
5-Minute Check on Chapter 45-Minute Check on Chapter 45-Minute Check on Chapter 45-Minute Check on Chapter 4 Transparency 5-1
Refer to the figure.1. Classify the triangle as scalene, isosceles, or equilateral.
2. Find x if mA = 10x + 15, mB = 8x – 18, andmC = 12x + 3.
3. Name the corresponding congruent angles if RST UVW.
4. Name the corresponding congruent sides if LMN OPQ.
5. Find y if DEF is an equilateral triangle and mF = 8y + 4.
6. What is the slope of a line that contains (–2, 5) and (1, 3)?Standardized Test Practice:
A CB D–3/2–2/3 2/3 3/2
5-Minute Check on Chapter 45-Minute Check on Chapter 45-Minute Check on Chapter 45-Minute Check on Chapter 4 Transparency 5-1
Refer to the figure.1. Classify the triangle as scalene, isosceles, or equilateral.
isosceles
2. Find x if mA = 10x + 15, mB = 8x – 18, andmC = 12x + 3. 6
3. Name the corresponding congruent angles if RST UVW. R U; S V; T W
4. Name the corresponding congruent sides if LMN OPQ.LM OP; MN PQ; LN OQ
5. Find y if DEF is an equilateral triangle and mF = 8y + 4. 7
6. What is the slope of a line that contains (–2, 5) and (1, 3)?Standardized Test Practice:
A CB D–3/2–2/3 2/3 3/2
Objectives
• Identify and use perpendicular bisectors and angle bisectors in triangles
• Identify and use medians and altitudes in triangles
Vocabulary• Concurrent lines – three or more lines that intersect at
a common point
• Point of concurrency – the intersection point of three or more lines
• Perpendicular bisector – passes through the midpoint of the segment (triangle side) and is perpendicular to the segment
• Median – segment whose endpoints are a vertex of a triangle and the midpoint of the side opposite the vertex
• Altitude – a segment from a vertex to the line containing the opposite side and perpendicular to the line containing that side
Vocabulary• Circumcenter – the point of concurrency of the
perpendicular bisectors of a triangle; the center of the largest circle that contains the triangle’s vertices
• Centroid – the point of concurrency for the medians of a triangle; point of balance for any triangle
• Incenter – the point of concurrency for the angle bisectors of a triangle; center of the largest circle that can be drawn inside the triangle
• Orthocenter – intersection point of the altitudes of a triangle; no special significance
Theorems• Theorem 5.1 – Any point on the perpendicular bisector of a
segment is equidistant from the endpoints of the segment.• Theorem 5.2 – Any point equidistant from the endpoints of
the segments lies on the perpendicular bisector of a segment.
• Theorem 5.3, Circumcenter Theorem – The circumcenter of a triangle is equidistant from the vertices of the triangle.
• Theorem 5.4 – Any point on the angle bisector is equidistant from the sides of the triangle.
• Theorem 5.5 – Any point equidistant from the sides of an angle lies on the angle bisector.
• Theorem 5.6, Incenter Theorem – The incenter of a triangle is equidistant from each side of the triangle.
• Theorem 5.7, Centroid Theorem – The centroid of a triangle is located two thirds of the distance from a vertex to the midpoint of the side opposite the vertex on a median.
Triangles – Perpendicular Bisectors
C
Circumcenter
Note: from Circumcenter Theorem: AP = BP = CP
Midpoint of AB
Midpoint of BC
Midpoint of ACZ
Y
XP
B
A
Circumcenter is equidistant from the vertices
Triangles – Angle Bisectors
B
C
A
Incenter
Note: from Incenter Theorem: QX = QY = QZ
Z
X
Y
Q
Incenter is equidistant from the sides
Triangles – Medians
C
Midpoint of AB
Midpoint of BC
Midpoint of AC
Centroid
Medianfrom B
M
Z
Note: from Centroid theorem BM = 2/3 BZ
Y
X
Centroid is the point of balance in any triangle
B
A
Triangles – Altitudes
C
X
Z
Y
Orthocenter
Altitudefrom B
Orthocenter has no special significance for us
B
ANote: Altitude is the shortest distance
from a vertex to the line opposite it
Special Segments in Triangles
Name TypePoint of
ConcurrencyCenter Special
QualityFrom / To
Perpendicular
bisector
Line, segment or
ray
Circumcenter
Equidistantfrom vertices
Nonemidpoint of
segment
Angle bisector
Line, segment or
ray
IncenterEquidistantfrom sides
Vertexnone
Median segment CentroidCenter ofGravity
Vertexmidpoint of
segment
Altitude segment Orthocenter none Vertexnone
Location of Point of Concurrency
Name Point of Concurrency Triangle ClassificationAcute Right Obtuse
Perpendicular bisector Circumcenter Inside hypotenuse Outside
Angle bisector Incenter Inside Inside Inside
Median Centroid Inside Inside Inside
Altitude Orthocenter Inside Vertex - 90 Outside
Given:
Find:
Proof:
Statements Reasons
1. 1. Given
2. 2. Angle Sum Theorem3. 3. Substitution4. 4. Subtraction Property 5. 5. Definition of angle
bisector6. 6. Angle Sum Theorem7. 7. Substitution8. 8. Subtraction Property
mDGE
Find:
Given:
.
Proof:
Statements. Reasons
1. Given
2. Angle Sum Theorem3. Substitution4. Subtraction Property 5. Definition of angle bisector6. Angle Sum Theorem7. Substitution8. Subtraction Property
1.
2.3.4.5.
6.7.8.
1. Given
mADC
ALGEBRA Points U, V, and W are the midpoints of respectively. Find a, b, and c.
Find a.
Segment Addition Postulate
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 14.8 from each side.
Divide each side by 4.
Find b.
Segment Addition Postulate
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 6b from each side.
Divide each side by 3.
Subtract 6 from each side.
Find c.
Segment Addition Postulate
Centroid Theorem
Substitution
Multiply each side by 3 and simplify.
Subtract 30.4 from each side.
Divide each side by 10.
Answer:
ALGEBRA Points T, H, and G are the midpoints of respectively. Find w, x, and y.
Answer:
Summary & Homework
• Summary:– Perpendicular bisectors, angle bisectors, medians
and altitudes of a triangle are all special segments in triangles
– Perpendiculars and altitudes form right angles– Perpendiculars and medians go to midpoints– Angle bisector cuts angle in half
• Homework: – Day 1: pg 245: 46-49– Day 2: pg 245: 51-54
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