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Geometry - Exam 2 (Proofs)

Terms in this set (18)

Pasch's Theorem
Assume ℓ intersects AB, but not BC. Then A and B are on opposite sides of ℓ and B and C are on the same side of ℓ. It follows that A and C are on opposite sides of ℓ which means that ℓ intersects AC. (Here the Plane Sep. Postulate was used.)
Isosceles Triangle Theorem
By SAS, ∆ABC≅∆ACB (A↔A, B↔C). In follows that ∠B≅∠C.
Exterior Angle Theorem
Let M be the midpoint of AC and B*M*X such that BM≅MX. By SAS, ∆ABM≅∆CXM and hence ∠A≅∠ACX. It follows that m∠A < m∠ACD if we show that CA*CX*CD. This can be done (in several steps) using B*M*X and B*C*D via Plane Separation and the Interior Lemma.
To show that m∠B < m∠ACD we can reduce to the previous case by using the fact that ∠ACD≅∠BCE, where A*C*E. (vertical angle theorem)
Suppose ∆ABC is a triangle. Then (m∠A = α) + (m∠B = β) <180°. (NUMBER 5)
Let δ = m∠CBD, where A*B*D. By the Linear Pair Thm, α<δ. It follows that α<180°-β and hence α+β<180°.
Scalene Inequality
"→": Assume AC>BC. Let D∈AC such that CD≅CB. By the Isosc. Triangle Thm., ∠CBD≅CDB. Since A*D*C, it follows that BA*BD*BC and hence m∠CBA > m∠CBD. By the Exterior Angle Thm., m∠CDB > m∠DAB = m∠CAB. Putting everything together, we find
m∠CBA > m∠CBD = m∠CDB > m∠CAB
"←": Assume that m∠B > m∠A. By trichotomy, either AC<BC, AC=BC, or AC>BC. If AC<BC, then m∠A=m∠B by the Isosc. Triangle Thm., which is again a contradiction. Thus: AC>BC.
Suppose ∆ABC is a triangle such that m∠C = 90°. Then AB>BC and AB>AC. (NUMBER 7)
Set α=m∠A, β=m∠B, γ=m∠C=90°. By 5 (m∠A+m∠B<180°), α+γ < 180° and β+γ < 180° which implies (since γ=90°) that α<γ and β<γ. By the Scalene Inequality, AB>BC and AB>AC.
Triangle Inequality
Let D∈AB(ray) such that A*B*D and BD=BC. Then AB+BC=AD. By the Isosc Triangle Thm., ∠D≅∠BCD. Since A*B*D, we have CA*CB*CD and hence m∠ACD > m∠BCD = m∠ADC. Thus, by the Scalene Inequality, AB+BC=AD > AC.
Suppose l is a line and A is a point not on ℓ. Then there exists a point F on l such that AF ⊥ℓ (NUMBER 9)
Pick B,C,D∈ℓ, B*C*D. Construct r(ray) on the other side of ℓ from A, forming angle ≅∠ACD with CD(ray). Let A'∈r(ray) such that CA'≅CA. AA' intersects ℓ in some point F. By construction ∠A'CD≅∠ACD and ∠A'CB≅∠ACB. If F≠C it follows that ∆AFC≅∆A'FC by SAS. Thus, if F≠C, then ∠AFC and ∠A'FC form a linear pair of congruent angles and hence m∠AFC=90°. If F=C, then ∠AFD and ∠A'FD form a linear pair of congruent angles and hence m∠AFD=90°. In either case, AF⊥ℓ
The point F from 9 is unique
Suppose there exists another point F'≠F. Then ∆AFF' is a triangle such that m∠F+m∠F' = 180°. This contradicts the inequality from 5.
AF<AX for all points X on ℓ distinct from F.
Assume X∈ℓ and X≠F. Then ∆AFX is a right triangle with hypotenuse AX. By 7, AX>AF.
Alternate Interior Angle Thm
CONTRAPOSITIVE PROOF:
Assume ℓ and m are not parallel. Then ℓ and m intersect in some point C. By the Ext. Angle Thm., m∠1 > m∠4 and m∠3 > m∠2. So neither pair of alt. interior angles is congruent.
Suppose ℓ is a line and A is a point not on ℓ. Then there exists a line m such that A∈m and m‖ℓ.
Let F be the foot of the perpendicular from A to ℓ. By the Angle Construction Thm ∃ ray r that forms a right angle with AF(ray). Let m=r(line). By the Alt Int Angle Thm, m‖ℓ.
EPP HOLDS:
If ℓ and m are two parallel lines cut by transversal, then both pairs of alt. interior angles are congruent.
Assume ℓ‖m. Let ∠CAB and ∠DBA be a pair of alt. int. angles (with C∈ℓ and D∈m.) By the Angle Construction Thm, ∃ ray r=BE(ray) on the other side of t=AB(line) from C, forming an angle that is congruent to ∠CAB with BA(ray). By the Alt. Int. Angle Thm, r(line)‖ℓ. By EPP, r(line)=m and hence ∠ABD=∠ABE. It follows that ∠CAB≅∠ABD.
EPP HOLDS:
Suppose ∆ABC is a triangle. then m∠A+m∠B+m∠C=180°.
Let m be the (unique) line such that C∈m and m‖AB(line). Pick D,E∈m such that D*C*E, D is on the same side of BC(line) as A, and E is on the same side of AC(line) as B. Then CD*CA*CB and CA*CB*CE, which implies that m∠DCA+m∠ACB+m∠BCE=180°. By the Converse of the Alt Int Angle Thm, ∠DCA≅∠CAB and ∠BCE≅∠CBA. It follows that m∠CAB+m∠ACB+m∠CBA=180°.
EPP HOLDS:
Suppose ℓ and m are distinct lines and t is a transversal for ℓ and m as follows: *see drawing for 19 in notestes** If m∠2+m∠4 < 180°, then ℓ and m intersect on the same side as the angles ∠2 and ∠4.
Assume m∠2+m∠4<180°. Then ℓ and m are not parallel since ℓ‖m would imply m∠2+m∠4=180° by the Converse to the Consecutive Angle Thm. Let C be the intersection point. Suppose C lies on the side of ∠1 and ∠3. Then ∠1 and ∠3 are two angles of triangle ∆ABC and hence m∠1+m∠3<180° by 5. But then
m∠2+m∠4=(180°-m∠1)+(180°-m∠3)
=360°-(m∠1+m∠3)>180°.
This contradicts m∠2+m∠4<180°.
EPP HOLDS:
Given: segment AB and positive real numbers α,β,γ such that α+β+γ=180°
Then: ∃ triangle ∆ABC such that m∠A=α, m∠B=β, and m∠C=γ.
Construct rays on the same side of AB(line) such that angles of measures α and β are formed with AB(ray) and BA(ray), respectively. Since α+β<180°, it follows by Euclid's Fifth that the two rays meet at a point C, forming a triangle ∆ABC. By construction, m∠A=α and m∠B=β. Since α+β+γ=180° and α+β+m∠C=180°, we also have m∠C=γ.
EPP HOLDS:
Suppose ABCD is a parallelogram. Then AB≅CD and BC≅DA. (NUMBER 21)
Since every parallelogram is a convex quadrilateral, B and D are on opposite sides of AC(line). By the Converse of the Alt Int Angle Thm,
∠BAC≅∠DCA (since AB‖CD) and
∠BCA≅∠DAC (since BC‖DA).
By ASA, ∆ABC≅∆CDA and hence AB≅CD and BC≅DA.
EPP HOLDS:
Suppose ABCD is a rectangle. Then AC≅BD.
By the Alt Int Angle Thm, ABCD is also a parallelogram. Thus, by 21, AB≅CD (and BC≅DA). Now, by SAS, ∆ABC≅∆BAD and hence AC≅BD.
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