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Home , Transversal (geometry)

Homework 16 Hints and Answers

Hint Section 4.7 #2: EEP implies #1: Assume t a that intersects l but not l', and find a contradiction. #1 implies EEP: Assume two lines through P and use #1 to show that one of them has to cross l.

Hint Section 4.7 #3 EEP implies #2: Let P be the of intersection between l and t. Use double construction to construct the desired perpendicular to t, and then show this line equals l'. #2 implies EEP: Drop a perpendicular from P to l and assume two lines through P, and show these lines are equal.

Hint Section 4.7 #4 EEP implies #3: Set up the picture suggested by the hypothesis. Identify lots of right and use those to conclude that m || n. #3 implies EEP: Create one parallel line through double perpendicular construction, and assume another parallel line through P. Drop a perpendicular from the foot on l to the second parallel line. Then use #3 to draw conclusions.

Hint Section 4.7 #5 EEP implies #4: Look at the two cases when (1) l || n and (2) l not parallel to n. #4 implies EEP: Assume two parallel lines through P and use #4 to draw conclusions.

2. Euclidean implies Proclus's Axiom: Let l and l' be parallel lines. Let t ≠ l be a line such that t intersects l. Suppose that t does not intersect l'. So t || l'. Let P be the intersection of l and t. Then there are two distinct lines through P that are parallel to l', which contradicts the Euclidean Parallel Postulate. So t crosses l'. Assume Proclus's Axiom. Let l be a line and P an external point of l. Suppose that m is parallel to l and passes through P. Let n be a line that passes through P and is distinct from m. Then since n intersects m, n also intersects l by Proclus's Axiom. So there is only one parallel line through P that is parallel to l.

3. Assume the Euclidean Parallel Postulate. Let l and l' be lines, l || l', and t a transversal such that t ⊥ l. Let P be the point at the intersection of l and t. Since t ⊥ l, then t ≠ l. And since there is only one line through P that is parallel to l', it follows that t is not parallel to l', so t must intersect l'. Let Q be the point of intersection between l' and t. By definition of perpendicular lines, the angles at the intersection of l and t are right angles. By the Converse to the Alternate Interior Angles Theorem, the angles at the intersection of t and l' are also right angles. Thus, by definition of perpendicular lines, l' ⊥ t. Assume Theorem 4.7.3 #2. Let l be a line and P a point external to l. Let t be the perpendicular dropped from P to l. Suppose that m and m' are two lines through P that are parallel to l. Then t ⊥ m and t ⊥ m'. Then the angles at the intersection of t and m are right angles, as are the angles at the intersection of t and m' (Def of Perp Lines). Then m = m' (Protractor Postulate). So there is only one line through P that is parallel to l.

4. Assume the Euclidean Parallel Postulate. Let l, m, n, and k be lines such that k || l, m ⊥ k, n ⊥ l. Note that k ≠ l, because the two lines are parallel, so they do not intersect. m l intersects l because there is only one unique line that passes through the intersection of k and m that is parallel to l. ∠1 Furthermore, the angles at the intersection of k and m must k be right angles (Def of Perpendicular). Likewise, n intersects k and forms right angles at the intersection. By the m Converse of the Alternate Interior Angles Theorem, ∠1 is a n right . By the Corresponding Angles Theorem, if the point of intersection between l and m and the point of intersection between l and n are different points, then m || n. If the points of intersection are equal, then by the uniqueness of n , m = n. P Now assume Theorem 4.7.3 #3. Let l be a line and P an n' external point. Let m be the perpendicular dropped for P to l, and m m' n be the line through P such that n ⊥ m and l || n (Double l Perpendicular Construction). Let n' be another line through P Q that is parallel to l. Let the foot of m on l be the point Q. Drop the perpendicular m' from Q to n'. Then n' ⊥ m', m ⊥ l, and l || n'. Then m and m' are parallel or equal. But m and m' intersect at Q, so m = m'. Thus, m ⊥ n', which implies n = n' (Protractor Postulate).

5. Assume the Euclidean Parallel Postulate. Let l, m, and n be lines such that l || m and m || n. If l and n never intersect, then l || n. If l and n intersect at a point P, then because m || l and m || n, P is not on m. By the Euclidean Parallel Postulate, there is only one line through P that is parallel to m. Therefore, l = n. Assume transitivity of parallelism. Let l be a line and P a point external to l. Let m and n be two lines through P that are parallel to l. Then m || n or m = n. Since m and n intersect at P, m is not parallel to n. So m = n.