Unit 5 Test Study Guide Relationships In Triangles [verified] -

Thus, the (intersection of perpendicular bisectors) is equidistant from all three vertices. That distance is the circumradius.

The compares two triangles that have two sets of congruent sides. unit 5 test study guide relationships in triangles

| Point of Concurrency | Formed by... | Key Property | Location in Acute Triangle | Location in Right Triangle | Location in Obtuse Triangle | |---------------------|--------------|--------------|----------------------------|----------------------------|-----------------------------| | | Perpendicular bisectors | Equidistant from all vertices (center of circumscribed circle) | Inside | At midpoint of hypotenuse | Outside | | Incenter | Angle bisectors | Equidistant from all sides (center of inscribed circle) | Inside | Inside | Inside | | Centroid | Medians | 2/3 of the distance from vertex to opposite side (center of mass) | Inside | Inside | Inside | | Orthocenter | Altitudes | No single distance property, but important for proofs | Inside | At the right angle vertex | Outside | | Point of Concurrency | Formed by

If ( G ) is centroid, ( AG : GD = 2 : 1 ) (where ( D ) is midpoint of ( BC )). One remote interior angle is 50°

An exterior angle of a triangle is 120°. One remote interior angle is 50°. Find the other remote interior angle and the adjacent interior angle. Solution: Other remote = 120° – 50° = 70°. Adjacent interior = 180° – 120° = 60°.

Triangles ( ABC ) and ( DEF ) have ( AB=DE, AC=DF ), ( \angle A=80^\circ, \angle D=60^\circ ). Compare ( BC ) and ( EF ).

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