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ASA- Angle Side Angle Used to prove triangle congruence: if two angles and the included side of two triangles are congruent, then the triangles are congruent. K L Given: KL and NO are parallel; M bisects KO. Prove: M KLM ≡ Statements Reasons KL and NO are parallel; M bisects KO. KML ≡ OMN ∟MKL ≡ ∟MON KM ≡ MO Given KLM ≡ N O ONM ONM Vertical Angles Alt. Interior Definition of bisect ASA Angle Angle Side -Two triangles can be proven to be congruent if two angles and the not included side are congruent. Statements Reasons DE=FG; DA ll EC; <B and <E are right angles Given <A = 90° <C = 90° Definition right angle <A = <C Transitive Property EF = EF Reflexive Property DF = GE Overlapping Segments <D = <E Corresponding Angles Postulate ABC = DEF AAS • Given: DE =FG ; DA ll EC; <B and <E are right angles • Prove: ABC = DEF A C B D E F G Side Angle Side • Given: AB = BC, AD = EC • Prove: ABE = CBD Statements Reasons AB = BC; AD = EC Given AB = CB Segment Addition <B = <B Reflexive Property ABE = ___CBD SAS B D E F A C Hypotenuse-Leg Statements Reasons <1 and <2 are Given right angles; AB = CB <1 = 90° <2 = 90° Definition right angle <1 = <2 Transitive Property BD = BD Reflexive Property HL ADB = ___CDB • Given: <1 and <2 are right angles; AB = CB • Prove: ADB = CDB D 1 A 2 B C Side Side Side Theorem Statements Reasons <1= <2, <3= <4 Given BF=BF Reflexive ABF= CBF ASA AB=BC CPCTC AF=CF CPCTC ABC is isosceles Def of isosceles BD – angle bisector Def- angle bisector BDperpendicular bisector Angle bisector of the vertex angle of an isos. triangle is a perpendicular bisector of the base AD=CD Def of perpendicular bisector FD=FD reflexive AFD= CFD SSS • Given: <1= <2, <3= <4 • Prove: AFD= CFD B F A D C Base Angle Theorem Statements AC=BC DC- angle bisector <ACD=<BCD CD=CD ACD= BCD <A=<B proofs Reasons Given construction • Given: AC=BC • Prove: <A=<B Def- angle bisector Reflexive SAS C A CPCTC D B A square is a rhombus Theorem Statements Reasons • Given: ABCD is a square Prove: ABCD is a rhombus ABCD is a square Given AB=BC=CD=DA Definition of a square ABCD is a rhombus Definition of a Rhombus A D B C If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram Statements Reasons BD bisects AC Given BE=ED, AE=EC Definition of a bisector <AEB=<DEC Vertical angles AEB= CED, AED= CEB SAS <ECD=<EAB, <ECB=<EAD CPCTC A B E D AB parallel to CD, Converse of alt. BC parallel to AD int. angles ABCD is a parallelogram • Given: BD bisects AC • Prove: ABCD is a parallelogram. definition C If one pair of adjacent sides of a parallelogram are congruent, then the parallelogram is a rhombus. Statements Reasons ABCD is a parallelogram, AB=BC Given AB=CD, BC=AD Opposite sides of a parallelogram are congruent CD=AB=BC=AD transitive ABCD is a rhombus definition • Given: ABCD is a parallelogram, AB=BC • Prove: ABCD is a rhombus. A D B C If the diagonals of a parallelogram bisect the angles of the parallelogram, then the parallelogram is a rhombus. Statements Reasons ABCD is a parallelogram, BD bisects <ADC and <ABC, AC bisects <BAD and <BCD Given <BAE=<DAE, <ADE=<CDE, <ABE=<CBE, <BCE=<DCE Definition of an angle bisector <EAD=<ECB Alt. int. angles <EAB=<ECB transitive BE=BE reflexive ABE= CBE AAS AB=BC CPCTC ABCD is a rhombus 1 pair of sides of a parallelogram are congruent. • Given: ABCD is a parallelogram, BD bisects <ADC and <ABC, AC bisects <BAD and <BCD. • Prove: ABCD is a rhombus. B C E A D If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus: Statements Reasons ABCD is a parallelogram, BD perpendicular to AC. Given BE=BE reflexive • Given: ABCD is a parallelogram, BD perpendicular to AC. • Prove: ABCD is a rhombus. B AE=EC Diagonals of a parallelogram bisect <BEC=90, <BEA=90 Definition of perpendicular <BEC=<BEA transitive ABE= CBE A E SAS AB=BC CPCTC ABCD is a rhombus 1 pair of adjacent sides of a parallelogram are congruent. D C