Transcript 5.1 Bisectors, Medians, Altitudes

Warm-ups
5.1 Bisectors, Medians, Altitudes
Term
Perpendicular bisector of a triangle – line segment or ray that
passes through the midpoint of the side and is perpendicular to
that side
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 a segment lies of
the perpendicular bisector of the segment
Concurrent Lines- when three or more lines intersect at a
common point
Point of Currency – point of intersection of concurrent lines
Circumcenter –point of currency of the perpendicular bisectors of
a triangle
Theorem 5.3- Circumcenter Theorem
The circumcenter of a triangle is equidistant from the vertices of
the triangle.
Incenter – the point of currency of the angle
bisectors of a 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.
Median- a segment whose endpoints are a vertex of a triangle
and the midpoint of the side opposite the vertex. (3) (It is also
the “bisector” of the side of the triangle)
Centroid – the point of concurrency for the medians of a triangle.
It is the point of balance.
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
the median.
Altitude – a segment from a vertex to the line containing
the opposite side and perpendicular to the line
containing that side. (3)
Orthocenter – the intersection point of the altitudes
Given:
Prove:
Proof:
Statements
Reasons
1.
1. Given
2.
3.
4.
5.
2. Angle Sum Theorem
3. Substitution
4. Subtraction Property
5. Definition of angle
bisector
6.
7.
8.
6. Angle Sum Theorem
7. Substitution
8. Subtraction Property
Given:
.
Prove:
Proof:
Statements.
Reasons
1.
1. Given
2.
3.
4.
5.
2. Angle Sum Theorem
3. Substitution
4. Subtraction Property
5. Definition of angle
bisector
6. Angle Sum Theorem
7. Substitution
8. Subtraction Property
6.
7.
8.
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.
Subtract 6 from each side.
Divide each side by 3.
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:
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.
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.
Next, find an equation of the altitude from I to
The
slope of
so the slope of an altitude is –6.
Point-slope form
Distributive Property
Subtract 3 from each side.
Then, solve a system of equations to find the point
of intersection of the altitudes.
Equation of altitude from J
Substitution,
Multiply each side by 5.
Add 105 to each side.
Add 4x to each side.
Divide each side
by –26.
Replace x with
y-coordinate.
in one of the equations to find the
Rename as improper fractions.
Multiply and simplify.
Answer:
The coordinates of the orthocenter of
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)