# By: Aaron Friedman-Heiman

## Presentation on theme: "By: Aaron Friedman-Heiman"— Presentation transcript:

By: Aaron Friedman-Heiman
Chapter 4 Proofs By: Aaron Friedman-Heiman and David Oliver

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. Given: KL and NO are parallel; M bisects KO. Prove: KLM ≡ ONM Statements Reasons KL and NO are parallel; M bisects KO. KML ≡ OMN ∟MKL ≡ ∟MON KM ≡ MO KLM ≡ ONM Given 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 E F G D

Side Angle Side Statements Reasons AB = BC; AD = EC Given AB = CB
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 right angles; AB = CB Given <1 = 90° <2 = 90° Definition right angle <1 = <2 Transitive Property BD = BD Reflexive Property ADB = ___CDB HL Given: <1 and <2 are right angles; AB = CB Prove: ADB = CDB D A C B

Side Side Side Theorem Given: <1= <2, <3= <4
Statements Reasons <1= <2, <3= <4 Given BF=BF Reflexive ABF= CBF ASA AB=BC CPCTC AF=CF ABC is isosceles Def of isosceles BD – angle bisector Def- angle bisector BD- perpendicular 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 6

Base Angle Theorem Given: AC=BC Prove: <A=<B Statements Reasons
DC- angle bisector construction <ACD=<BCD Def- angle bisector CD=CD Reflexive ACD= BCD SAS <A=<B CPCTC Given: AC=BC Prove: <A=<B C A D B proofs 7

A square is a rhombus Theorem
Given: ABCD is a square Prove: ABCD is a rhombus Statements Reasons ABCD is a square Given AB=BC=CD=DA Definition of a square ABCD is a rhombus Definition of a Rhombus A B D C 8

Prove: ABCD is a parallelogram.
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 AB parallel to CD, BC parallel to AD Converse of alt. int. angles ABCD is a parallelogram definition Given: BD bisects AC Prove: ABCD is a parallelogram. A B E D C 9

Given: ABCD is a parallelogram, AB=BC
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 B D C 10

Prove: ABCD is a rhombus.
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 11

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 AE=EC Diagonals of a parallelogram bisect <BEC=90, <BEA=90 Definition of perpendicular <BEC=<BEA transitive ABE= CBE SAS AB=BC CPCTC ABCD is a rhombus 1 pair of adjacent sides of a parallelogram are congruent. Given: ABCD is a parallelogram, BD perpendicular to AC. Prove: ABCD is a rhombus. B E A C D 12