Download presentation

Presentation is loading. Please wait.

1
**By: Aaron Friedman-Heiman**

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

2
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

3
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

4
**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

5
**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

6
**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

7
**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

8
**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

9
**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

10
**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

11
**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

12
**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

Similar presentations

OK

Proving Triangles Congruent. Steps for Proving Triangles Congruent 1.Mark the Given. 2.Mark … reflexive sides, vertical angles, alternate interior angles,

Proving Triangles Congruent. Steps for Proving Triangles Congruent 1.Mark the Given. 2.Mark … reflexive sides, vertical angles, alternate interior angles,

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on review of literature in research Ppt on live line maintenance tools Ppt on water scarcity graph Ppt on obesity management protocols Ppt on australian continent history Ppt on digital media piracy control Download ppt on law and social justice Ppt on nuclear power plants in india Ppt on high level languages Ppt on condition based maintenance navy