perpendicular bisector- a line, segment, or ray that passes through the midpoint of the side and is...

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)