Geometry Conjectures Flashcards


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created 6 months ago by Gdufie4
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1

C-14 Median Concurrency

The three medians of a triangle are concurrent.

The point of Concurrency is called the centroid.

2

C-11 Altitude Concurrency

The three altitudes of a triangle are concurrent.

The point of concurrency is called the Orthocenter

3

C-10 Perpendicular Bisector Concurrent

The three Perpendicular Bisectors are concurrent.

The point of concurrency is called the circumcenter.

4

C-9: Angle Bisector Concurrency

The three angle bisectors of a
triangle meet at a point.
-or-
The three angle bisectors of a
triangle are concurrent.
C-9: Angle Bisector Concurrency
The point of concurrency is called the incenter

5

C-16: Center of Gravity

The centroid of a triangle is the
center of gravity of the triangular
region.

6

C-15: Centroid Conjecture

The centroid divides the median into
2 segments so that the distance
from the vertex to the centroid is
twice the distance from the centroid to the midpoint.

7

Orthocenter Conjectures

The orthocenter of a right triangle is
on the vertex of the right angle.

Altitudes
The orthocenter of an obtuse
triangle is outside of the triangle
Orthocenter Conjectures

8

C-12: Circumcenter Conjecture

The circumcenter is equidistant from
the vertices of a triangle.

Perpendicular Bisector
Other info:
a. The circumcenter is outside of an obtuse
triangle
b. The circumcenter is on the midpoint of the
hypotenuse of a right triangle
c. The circumcenter is the center of the circle that
circumscribes the triangle

9

C-13: Incenter Conjecture

The incenter is equidistant from the sides of a triangle.

Angle Bisector

Other info:
a. The incenter is the center of the circle that is
inscribed inside the triangle
b. Don’t forget that the distance in question is the
shortest distance, perpendicular from the
incenter to the side of the triangle

10

Euler Line Conjecture

The circumcenter, centroid and
orthocenter are collinear.
For an isosceles triangle, the points
of concurrency all lie on the angle
bisector of the vertex angle.
For an equilateral triangle, all 4
points of concurrency are the same

11

Euler Segment Conjecture

The centroid divides the Euler
Segment into 2 parts so that the
smaller part is half of the larger part.