Transcript
Page 1: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

5.1 Bisector, Medians, and Altitudes

Page 2: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

A. Vocabulary:

1. A Median of a triangle is a segment from a vertex to the midpoint of the opposite side.

2. An Altitude of a triangle is the perpendicular segment from a vertex to

the line that contains the opposite side.

a. In an acute : b. In a right : c. In an obtuse :

Page 3: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

3. A Perpendicular Bisector of a segment is a line (or ray or segment) that is perpendicular to the segment at its midpoint.

4. Circumcenter is the point where the three perpendicular bisectors meet inside the triangle. (Point of intersection is know as the point of concurrency)

5. Centroid is the point of concurrency where the medians of the triangle meet.

6. Incenter is the point of concurrency where the angle bisectors of the triangle meet.

7. Orthocenter is the point of concurrency where the altitudes of the triangle meet.

Page 4: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

B. Theorem 5.1:If a point lies on the perpendicular bisector of a segment, then

the point is equidistant from the endpoints of the segment.

6x - 53x + 16

A

B CD

Example 1: Find x, if AD is a perpendicular bisector.

Page 5: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

C. Theorem 5.3: Circumcenter TheoremThe circumcenter of a triangle is equidistant from the

vertices of the triangle.

A

B

C

E

Example 2

Find x and y, if BE = 24,

AE = 3x – 6 and CE = 5y + 4

Page 6: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

D. Theorem 5.4:If a point lies on the bisector of an angle, then

the point is equidistant from the sides of the angle.

AB

C

D

Example 3

Find x, if AC is an angle bisector of DAB and DC = x + 4 and BC = 2x – 5.

Page 7: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

E. Theorem 5.7: Centroid TheoremThe distance from the vertex to the point of concurrency is equal to 2 times the distance from the point of concurrency

to the midpoint of the opposite side. Example 4

Point A is a centroid of ∆DEF. Find x, y, and z.

D

S

E

T

FU

A

SA = 4z

EA = y

TA = 2x – 5

FA = 4.6

UA = 2.9

DA = 6

Page 8: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

Example 5 Find x and IJ if HK is an altitude of HIJ. Is HK a perpendicular bisector? Explain.

I

J

K H (3x + 3)

x + 8

x – 9

Page 9: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

Example 6

Identify the following: a. An angle bisector b. A median c. A perpendicular bisector d. An altitude

A

G

F

D C B

E

Page 10: 5.1 Bisector, Medians, and Altitudes. B. Theorem 5.1: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from

Example 7 Line CD is the perpendicular bisector of both segment XY and ST, and CY = 16. Find each length.

a. CT b. TY c. SX d. CX e. MT f. ST g. DY h. XY

M S

C

T

16

X 7 D Y

12

5


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