1 st proof parallel postulate perpendicular postulate alternate
- Slides: 14
1 st proof parallel postulate perpendicular postulate alternate interior angles theorem 2 nd proof triangle sum theorem
B Given: ∆ABC <1 <2 Prove: <3 A C Statement Reason ∆ABC given <1 + <2 + <3 = 180º
B D E Given: ∆ABC <5 <6 <1 <2 A Prove: <1 + <2 + <3 = 180º <3 C Statement Reason ∆ABC Draw line DE through point B, parallel to line AC given parallel postulate
B D E Given: ∆ABC <5 <6 <1 <2 A Prove: <1 + <2 + <3 = 180º <3 C Statement Reason ∆ABC Draw line DE through point B, parallel to line AC <DBC + <6 = 180º given parallel postulate linear pair postulate
B D E Given: ∆ABC <5 <6 <1 <2 A Prove: <1 + <2 + <3 = 180º <3 C Statement Reason ∆ABC Draw line DE through point B, parallel to line AC <DBC + <6 = 180º <DBC = <1 + <5 given parallel postulate linear pair postulate angle addition postulate
B D E Given: ∆ABC <5 <6 <1 <2 A Prove: <1 + <2 + <3 = 180º <3 C Statement Reason ∆ABC Draw line DE through point B, parallel to line AC <DBC + <6 = 180º <DBC = <1 + <5 + <6 = 180º given parallel postulate linear pair postulate angle addition postulate substitution
B D E Given: ∆ABC <5 <6 <1 <2 A Prove: <1 + <2 + <3 = 180º <3 C Statement Reason ∆ABC Draw line DE through point B, parallel to line AC <DBC + <6 = 180º <DBC = <1 + <5 + <6 = 180º <2 = <5 given parallel postulate linear pair postulate angle addition postulate substitution alternate interior angle theorem
B D E Given: ∆ABC <5 <6 <1 <2 A Prove: <1 + <2 + <3 = 180º <3 C Statement Reason ∆ABC Draw line DE through point B, parallel to line AC <DBC + <6 = 180º <DBC = <1 + <5 + <6 = 180º <2 = <5 <3 = <6 given parallel postulate linear pair postulate angle addition postulate substitution alternate interior angle theorem
B D E Given: ∆ABC <5 <6 <1 <2 A Prove: <1 + <2 + <3 = 180º <3 C Statement Reason ∆ABC Draw line DE through point B, parallel to line AC <DBC + <6 = 180º <DBC = <1 + <5 + <6 = 180º <2 = <5 <3 = <6 <1 + <2 + <3 = 180º given parallel postulate linear pair postulate angle addition postulate substitution alternate interior angle theorem substitution
B Given: ∆ABC <1 <2 Prove: <3 <4 A C D Statement Reason ∆ABC given <1 + <2 = <4
B Given: ∆ABC <1 <2 A Prove: <1 + <2 = <4 <3 <4 C D Statement Reason ∆ABC <1 + <2 + <3 = 180º given triangle sum theorem
B Given: ∆ABC <1 <2 A Prove: <1 + <2 = <4 <3 <4 C D Statement Reason ∆ABC <1 + <2 + <3 = 180º <3 + <4 = 180º given triangle sum theorem linear pair postulate
B Given: ∆ABC <1 <2 A Prove: <1 + <2 = <4 <3 <4 C D Statement Reason ∆ABC <1 + <2 + <3 = 180º <3 + <4 = 180º <1 + <2 + <3 = <3 + <4 given triangle sum theorem linear pair postulate substitution
B Given: ∆ABC <1 <2 A Prove: <1 + <2 = <4 <3 <4 C D Statement Reason ∆ABC <1 + <2 + <3 = 180º <3 + <4 = 180º <1 + <2 + <3 = <3 + <4 <1 + <2 = <4 given triangle sum theorem linear pair postulate substitution subtraction property
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