5.1 bisectors, medians, and altitudes. objectives identify and use ┴ bisectors and bisectors in...
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5.1 Bisectors, Medians,
and Altitudes
Objectives
Identify and use ┴ bisectors and bisectors in ∆s
Identify and use medians and altitudes in ∆s
Perpendicular Bisector
A ┴ bisector of a ∆ is
a line, segment, or
ray that passes
through the midpoint
of one of the sides of
the ∆ at a 90° .
Side AB
perpendicular bisector
PA B
C
┴ Bisector Theorems
Theorem 5.1 – Any point on the ┴ bisector of a segment is equidistant from the endpoints of the segment.
Theorem 5.2 – Any point equidistant from the endpoints of a segment lies on the ┴ bisector of the segment.
┴ Bisector Theorems (continued)
Basically, if CP is the perpendicular bisector of AB, then…
Side AB
CP is perpendicular bisector
PA B
C
CA ≅ CB.
┴ Bisector Theorems (continued)
Since there are three sides in a ∆, then there are three ┴ Bisectors in a ∆.
These three ┴ bisectors in a ∆ intersect at a common point called the circumcenter.
┴ Bisector Theorems (continued)
Theorem 5.3 (Circumcenter Theorem) The circumcenter of a ∆ is equidistant from the vertices of the ∆.
Notice, a circumcenter of a ∆ is the center of the circle we would draw if we connected all of the vertices with a circle on the outside (circumscribe the ∆).
circumcenter
Angle Bisectors of ∆s Another special bisector which we have already
studied is an bisector. As we have learned, an bisector divides an into two parts. In a ≅ ∆, an bisector divides one of the ∆s s into two ≅ s.(i.e. if AD is an bisector then BAD ≅ CAD)
B D C
Angle Bisectors of ∆s (continued)
Theorem 5.4 – Any point on an bisector is equidistant from the sides of the .
Theorem 5.5 – Any point equidistant from the sides of an lies on the bisector.
Angle Bisectors of ∆s (continued)
As with ┴ bisectors, there are three bisectors in any ∆. These three bisectors intersect at a common point we call the incenter.
incenter
Angle Bisectors of ∆s (continued)
Theorem 5.6 (Incenter Theorem) The incenter of a ∆ is equidistant from each side of the ∆.
Medians
A median is a segment whose endpoints are a vertex of a ∆ and the midpoint of the side opposite the vertex. Every ∆ has three medians.
These medians intersect at a common point called the centroid.
The centroid is the point of balance for a ∆.
Medians (continued)
Theorem 5.7 (Centroid Theorem)The centroid of a ∆ is located two thirds of the distance from a vertex to the midpoint of the side opposite the vertex on a median.
CENTROID
acute triangle
P
Altitudes
An altitude of a ∆ is a segment from a vertex to the line containing the opposite side and ┴ to the line containing that side. Every ∆ has three altitudes.
The intersection point of the altitudes of a ∆ is called the orthocenter.
Altitudes (continued)
G
F
D
E
B
A E
Altitude
Orthocenter
ALGEBRA: Points U, V, and W are the midpoints of respectively. Find a, b, and c.
Example 1:
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.
Example 1:
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.
Example 1:
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:
Example 1:
ALGEBRA: Points T, H, and G are the midpoints of respectively. Find w, x, and y.
Answer:
Your Turn:
COORDINATE GEOMETRY The vertices of HIJ are H(1, 2), I(–3, –3), and J(–5, 1). Find the coordinates of the orthocenter of HIJ.
Example 2:
Find an equation of the altitude from The slope of
so the slope of an altitude is
Point-slope form
Distributive Property
Add 1 to each side.
Example 2:
Point-slope form
Distributive Property
Subtract 3 from each side.
Next, find an equation of the altitude from I to The
slope of so the slope of an altitude is –6.
Example 2:
Equation of altitude from J
Multiply each side by 5.
Add 105 to each side.
Add 4x to each side.
Divide each side by –26.
Substitution,
Then, solve a system of equations to find the point of intersection of the altitudes.
Example 2:
Replace x with in one of the equations to find the y-coordinate.
Multiply and simplify.
Rename as improper fractions.
The coordinates of the orthocenter of Answer:
Example 2:
COORDINATE GEOMETRY The vertices of ABC are A(–2, 2), B(4, 4), and C(1, –2). Find the coordinates of the orthocenter of ABC.
Answer: (0, 1)
Your Turn:
Assignment
Geometry: Pg. 242 #6, 13 – 26
Pre - AP Geometry:Pg. 242 #6 – 9, 13 – 30