Helpful properties and postulates RReflexive property SSegment and Angle addition postulate AAddition Property of Equality SSubtraction Property.

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Presentation transcript:

Helpful properties and postulates RReflexive property SSegment and Angle addition postulate AAddition Property of Equality SSubtraction Property of Equality TTransitive and Substitution Property

Reflexive property When triangles share the same side, the side is congruent to itself. This is called the REFLEXIVE PROPERTY.

Segment Addition Postulate  A whole is equal to the sum of all its parts. The Postulate says that if point Q is between points P and R on a line, then PQ + QR = PR. NOTE: Notice there is a plus sign on only one side of the equation

Angle Addition Postulate AA whole is equal to the sum of all its parts.

Substitution OR Transitive Property of Equality  Transitive: If a = b and b = c, then a = c  Ex: Pig=bacon and bacon =fat, then pig=fat! Substitution : a quantity may be substituted for its equal in any expression If a statement a = b, is true and If the statement a + c = 180 is true then the statement b + c = 180 is also true Ex: Carmen= awesome math student and Pete = awesome math student, then Carmen= Pete! Hmmm… ??? Logical??? HA HA! In general: Transitive may be used when the order is the same as the above… Substitution may be used when any 2 quantities are equal! When in doubt, use substitution property!

Addition Property of Equality AAllows you to add congruent (equal) segments AAllows you to add congruent (equal) angles. NOTE: If there is a plus sign on both sides, then the reason is more likely to be Addition Property StatementsReasons AB = DE; BC = EF Given AB+BC=DE+EFAddition Property of Equality (added equal segments to both sides) AB+BC=AC; DE+EF=DF Segment Addition Postulate AC=DFSubstitution Property Given: AB=DE; BC=EF Prove: AC = DF

Subtraction Property of Equality AAllows you to subtract congruent (equal) segments AAllows you to subtract congruent (equal) angles. StatementsReasons AC = DF; BC = EF Given AC-BC=DF-EFSubtraction Property of Equality (subtracted equal segments from both sides) AC-BC=AB; DF-EF=DE Segment Subtraction Postulate AB=DESubstitution Property Given: AC=DF; BC=EF Prove: AB = DE