Upload
ruth-robbins
View
215
Download
0
Embed Size (px)
Citation preview
Perpendicular Bisector
Perpendicular Bisector- a 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 endpoint s of a
segment lies on the perpendicular bisector of the segment
Circumcenter
Concurrent lines- when 3 or more lines intersect at a common point
Point of Concurrency- the point that the concurrent lines intersect at
Circumcenter- the point of concurrency of the perpendicular bisectors of a triangle
Theorem 5-3 Circumcenter Theorem The circumcenter of a triangle is equidistant
from the vertices of a triangle
Point on Angle Bisectors
Theorem 5.4 Any point on the angle bisector is equidistant
from the sides of the angle Theorem 5.5
Any point equidistant from the sides of an angle lies on the angle bisector
Incenter- the point of concurrency of the angle bisectors of a triangle.
Theorem 5.6 Incenter Theorem The incenter of a triangle is equidistant from
each side of the triangle
Medians and Altitudes
Median- segment whose endpoints are a vertex of a triangle and the midpoint of the side opposite the vertex
Centroid- the point of concurrency for the medians of a 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 the median
Means and Altitude (con’t)
Altitude- segment from the vertex to the line containing the opposite side and perpendicular to the line containing that side
Orthocenter- the intersection point of the altitudes of a triangle
Given:
Prove:
Proof:
Statements Reasons
1. 1. Given
2. 2. Angle Sum Theorem
3. 3. Substitution
4. 4. Subtraction Property
5. 5. Definition of angle bisector
6. 6. Angle Sum Theorem
7. 7. Substitution
8. 8. Subtraction Property
Prove:
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
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:
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.
Point-slope form
Distributive Property
Subtract 3 from each side.
Next, find an equation of the altitude from I to Theslope of so the slope of an altitude is –6.
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.
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:
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)