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Birkoff’s Postulates Some Defined Terms Euclidean Parallel Postulate Existence of Perpendiculars Parallel Postulate Becomes a Theorem
Birkoff’s Postulates Some Defined Terms Euclidean Parallel Postulate Existence of Perpendiculars Parallel Postulate Becomes a Theorem
Postulate I: Postulate of Line Measure. A set of points fA, B...g on any line can be put into a 1: correspondence with the real numbers fa, b...g so that jb − aj = d(A, B) for all points A and B.
Postulate II: Point-Line Postulate. There is one and only one line, `, that contains any two given distinct points P and Q.
Postulate III: Postulate of Angle Measure. A set of rays fl, m, n...g through any point O can be put into 1:1 correspondence with the real numbers a(mod2Π) so that if A and B are points (not equal to O) of l and m, respectively, the difference am − al (mod2Π) of the numbers associated with the lines l and m is the measure of ∠AOB.
Postulate IV: Postulate of Similarity. Given two triangles ABC and A′B′C ′^ and some constant k > 0, d(A′, B′) = kd(A, B), d(A′, C ′) = kd(A, C ) and ∠B′A′C ′^ = ∠BAC , then d(B′, C ′) = kd(B, C ), ∠C ′B′A′^ = ∠CBA, and ∠A′C ′B′^ = ∠ACB
Birkoff’s Postulates Some Defined Terms Euclidean Parallel Postulate Existence of Perpendiculars Parallel Postulate Becomes a Theorem
If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.
Line
↔ BD and line
↔ AC will intersect if m∠CAD + m∠ADB < 180 ◦
Birkoff’s Postulates Some Defined Terms Euclidean Parallel Postulate Existence of Perpendiculars Parallel Postulate Becomes a Theorem
We will assume that given a line
↔ BC and a point A 6 ∈
↔ BC there exists a line passing through A and perpendicular to ↔ BC.
Birkoff’s Postulates Some Defined Terms Euclidean Parallel Postulate Existence of Perpendiculars Parallel Postulate Becomes a Theorem
1 Drop a perpendicular from B to C. 2 Find a point E on DF so that jjADAC jj = jjDEBC^ jj. This will make triangles 4 ADE and 4 ABC similar. 3 Thus m∠EAD = m∠BAC and protractor postulate implies that ↔ AE and
↔ AB are the same.
