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Triangle Segments
Special segments of a triangle, points of intersection, inequalities of triangle
| Vocab word | Definition |
|---|---|
| Angle bisector | splits an angle into congruent parts |
| Perpendicular bisector | splits a segment at its midpoint perpendicularly |
| Median | a segment that connects a vertex of a triangle to the midpoint of the opposite side |
| Altitude | a segment of a triangle that goes from a vertex perpendicularly to the opposite side |
| Midsegment | a segment in a triangle that connects two midpoints |
| Midpoint | a point in the middle of a segment that cuts it into two congruent parts |
| Orthocenter | the point of intersection of the altitudes |
| Circumcenter | the point of intersection of the perpendicular bisector |
| Incenter | the point of intersection of the angle bisectors |
| Centroid | the point of intersection of the medians |
| Center of inscribed circle | Incenter |
| Center of circumscribed circle | Circumcenter |
| 2/3 the distance from vertex to midpoint | Centroid |
| Equidistant from the sides of a triangle | Incenter |
| Equidistant from the vertices of the triangle | Circumcenter |
| Triangle Inequality Theorem | Two sides of a triangle must add up to be bigger than the third side |
| Largest side of a triangle | Opposite the largest angle |
| Smallest side of a triangle | Opposite the smalled angle |
| Largest angle of a triangle | Opposite the largest side |
| Smallest angle of a triangle | Opposite the smallest side |
| Hinge Theorem | If 2 sides of 2 triangles are congruent and the included angle of the first is bigger, then the third side of the first is bigger |
| Hinge Theorem Converse | If 2 sides of 2 triangles are congruent and the third side of the first is bigger, then the included angle of the first is bigger |
| Direct Proof | Using properties, definitions, postulates and theorems to prove something directly |
| Indirect Proof | Temporarily assuming the opposite of what you are trying to prove to reach a contradiction |
Created by:
jenkroesen
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