Geometry Honors C ONCURRENT L INES, M EDIANS & A LTITUDES.

CONCURRENT LINES, MEDIANS & ALTITUDES

Vocabulary

Concurrent Lines – when three or more lines intersect in one point.

Point of concurrency– the point at which 3 or more lines intersect.

Geogebra Demonstration of

Perpendicular Bisectors

Vocabulary

Circumcenter of the triangle– the point of concurrency of the perpendicular bisectors.

Circumcenter

The perpendicular bisectors of the sides of a triangle are concurrent at a point equidistant from the vertices.

Theorem

Since the vertices of the triangle are equidistant from the circumcenter, we can draw a circle around the triangle or circumscribe the triangle.

The center of the circle is the circumcenter of the triangle.

Geogebra Demonstration of Angle Bisectors

Vocabulary

Incenter of the triangle– the point of concurrency of the angle bisectors.

Incenter

Theorem

The bisectors of the angles of a triangle are concurrent at a point equidistant from the sides.

We can now inscribe a circle in the triangle since the incenter is equidistant from the sides.

The center of the circle is the incenter of the triangle.

Geogebra Demonstration of

Altitudes of Triangles

Vocabulary

Altitude of a Triangle– a perpendicular segment from a vertex to the line containing the side opposite to the vertex. Orthocenter

FYI

An altitude can be the side of the triangle.

Theorem

The lines that contain the altitudes of a triangle are concurrent.

Geogebra Demonstration of

Medians of Triangles

Vocabulary

Median of a Triangle– a segment whose endpoints are a vertex and the midpoint of the opposite side.

Centroid

FYI

The centroid is the center of gravity of the triangle. If you cut out the triangle and paper-fold to determine the centroid, you can balance the triangle on the tip of your pencil point.

Theorem

The medians of a triangle are concurrent at a point that is two thirds the distance from each vertex to the midpoint of the opposite side.

A

B

C

D

E

FG

Cut out your triangles.

Yellow Paper - Median•Draw an acute triangle.•Cut it out.•Determine the midpoint of each side by folding corner to corner. Then fold the triangle from the midpoint of each side to the vertex opposite it.•Ask Mrs. Slifko to see the model in my notebook if you need to .•Label the point of concurrency the centroid. Try to balance your triangle on the tip of your pencil.•Put any other theorems on the triangle.

Cut out your triangles.

Green Paper- Altitude•Draw an acute triangle.•Cut it out.•Fold the triangle so that you create an altitude extending from all three vertices.•Ask Mrs. Slifko to see the model in my notebook if you need to .•Label the point of concurrency the orthocenter. •Write any necessary information on the triangle.

We can use coordinate geometry to write equations of the various special segments of the triangles.

Example:

A(2,9)

B(-2,3) C(6,-1)

How can we find the slope of altitude AD?

D

We know that altitudes are at right angles to the

opposite sides…therefore, if we can find the slope of BC, we can use opposite

reciprocals to find the slope of AD.

Example:

A(2,9)

B(-2,3) C(6,-1)

How can we find the equation of altitude AD?

D

If you know a point on the altitude , namely

A(2,9),and the slope of AD, then you can use the point –slope formula to find the equation of altitude AD.

Example:

A(3,9)

B(-2,3) C(6,-1)

How can we find the slope of median AD?

D

Since we know that a median hits the midpoint,

we know that D is the midpoint of BC. Once you find the midpoint, you can use the slope formula to calculate the slope of AD.

Example:

A(3,9)

B(-2,3) C(6,-1)

How can we find the equation of median AD?

D

Since you know a point on the median line and a

slope, you can use point-slope formula to determine the equation of median AD.

Example:

A(6,9)

B(-2,3) C(12,-1)

How can we find the slope of the perpendicular bisector of BC?

D

Since we know that perpendicular bisectors

form right angles, we can determine the slope of BC

and then use opposite reciprocals to determine

the slope of the perpendicular bisector of

BC.