MAT 141 Midterm II - Key: Axioms, Proofs, and Interpretation of Hilbert Plane, Exercises of Analytical Geometry and Calculus

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MAT 141

Midterm II – Key

Name:

No notes. No calculators. All incidents of cheating or the appearance thereof will be brought to the attention of Student Judicial Affairs.

Total (out of 50):

1. (10 Points) State the 4 betweenness and the 6 congruence axioms.

B-1: If A ∗ B ∗ C , then A, B, C are distinct collinear points and C ∗ B ∗ A. B-2: Given distinct points B, D , there exist points A, C, E lying on

B

D such that A ∗ B ∗ D , B ∗ C ∗ D and B ∗ D ∗ E. B-3: If A, B, C are distinct collinear points, then one and only one of them is between the other two. B-4: For every line l and for any points A, B, C not lying on l : (i) If A, B are on the same side of l and B, C are on the same side of l , then A, C are on the same side of l. (ii) If A, B are on opposite sides of l and B, C are on opposite sides of l , then A, C are on the same side of l. C-1: If A, B are distinct points and A ′^ any point, then for every ray r emanating from A ′^ there is a unique point B ′ on r such that B ′^6 = A ′^ and AB ∼= A ′ B ′. C-2: If AB ∼= CD and AB ∼= EF , then CD ∼= EF. More- over, every segment is congruent to itself. C-3: If A ∗ B ∗ C , A ′^ ∗ B ′^ ∗ C ′, AB ∼= A ′ B ′^ and BC ∼= B ′ C ′, then AC ∼= A ′ C ′. C-4: Given any ^ BAC and given any ray

A ′ B ′^ emanating from a point A ′, there is a unique ray

A ′ C ′^ on a given side

of the line

A ′

B ′^ such that ^ B ′ A ′ C ′^ ∼= ^ BAC. C-5: If ^ A ∼= ^ B and ^ A ∼= ^ C , then ^ B ∼= ^ C. More- over, every angle is congruent to itself. C-6: If two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle in another triangle, then the two triangles are congruent to each other.

3. (10 Points) Use ruler and compass to circumscribe a circle around the given triangle.

4. (10 Points) Prove that, in a Hilbert plane, for every line l and every point P , there exists a line m incident to P perpendicular to l.

Proof: Let l be a line. Case 1: P is a point not on l. By Incidence Axiom I-2, there are two points A, B on l. By Congruence Axiom C-4 and Proposition 3.2, it is the case that on the opposite side of l from the ray − AP → , there exists a ray − AX −→ such that ^ XAB ∼= ^ P AB. By Congruence Axiom C-1, there is a point P ′^ on − AX −→ such that AP ′^ ∼= AP. The point P ′^ lies on the opposite side of l as P. By definition of opposite sides, P P ′^ is incident to l in a point Q. Subcase a: If Q = A , then, by definition of right angle, ← P − −→ P ′ (^) is perpendicular to l.

Subcase b: If Q 6 = A , then 4 P AQ ∼= 4 P ′ AQ , by Con- gruence Axiom C-6. In particular, ^ P QA ∼= ^ P ′ QA. By the definition of right angle, the line

P

P ′^ is perpen- dicular to l. This proves the statement in Case 1. Case 2: P is a point on l. By Proposition 2.3, there is a point C not on l. Proceed as in Case 1 to obtain a line m ′^ incident to C and perpendicular to l. Denote the point in which m ′^ is incident to l by Q. If Q = P , then m ′^ is the desired perpendicular. If Q 6 = P , then by Congruence Axiom C-4, there is a ray