Suggested problems - solutions

Transcription

1 Suggested problems - solutions Parallel lines Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, In particular, see section 4.1, pp The problems are all from section 4.1. Problems: 1, 2, 3, 5, 8, 10a, 13 #1: This is a two line proof, but as noted in the lecture, I d like to to keep track of the order you re going in with parallels and angles. Alternate interior angles congruent (or corresponding angles congruent) (or same side interior supplementary) imply parallel lines. is a version of Parallelism in Absolute Geometry. Parallel lines imply alternate interior angles congruent (or corresponding angles congruent) (or same side interior supplementary). is a version of the Parallel Postulate. Indicate which order if you re claiming the C, F, or Z properties. OK, now the two line proof. As always, put some points on the thing so you can refer to the segments and angles: Given: m 1 = m 2. Prove: m 3 = m 4. Since AB and CD are cut by transversal AC with corresponding angles 1 = 2 (given), we have AB CD (Property F, Parallelism in Absolute Geometry). Then, since AB CD and AB and CD are cut by transversal BD with corresponding angles 3 and 4, the corresponding angles must be congruent, and m 3 = m 4 (Property F, Parallel Postulate).

2 #2: Justify, justify, justify! And assume for this section that every problem is in Euclidean geometry, unless otherwise noted, or else you ll never get anywhere. Given: AB BD, ED BD, AC CE, m ACB = 31. Find: m 1, m 2, m m 1 = 180, so m 1 = 59 (Angle Sum Theorem). Careful for the next bit - yes, it is true that m 2 = but the thing you re thinking of is called the linear pair axiom, not the three angles sharing a line axiom. Do it right: m ACD + 31 = 180, so m ACD = 149 (Linear Pair Axiom). m DCE + m ECA = m ACD (Angle Addition Postulate). m = 149, so m 2 = m 3 = 180, so m 3 = 31 (Angle Sum Theorem).

3 #3: Given: Figure. Find: x, y, z z = 180, so z = 47 (Angle Sum Theorem). z + y = m ACB (Angle Addition Postulate), so 47 + y = 90 and y = 43. x + y + 90 = 180, so x = 180 and x = 47 (Angle Sum Theorem).

4 #5: Given: ABC equilateral, DE BC. Prove: ADF is isosceles. Since ABC is equilateral (given), it is also equiangular. [You know, we ve never explicitly proven this as a theorem. It s simply an application of the isosceles triangle theorem - an equilateral triangle is isosceles no matter you orient it - so all its base angles are equal.] m A = m B = m C m B + m D = 90 (angle sum is 180, with 90 taken up by the right angle - so corollary to the Angle Sum Theorem is that the acute angles of a right triangle are complementary). m C + m CF E = 90 (corollary to Angle Sum Theorem). m C + m CFE = m B + m D (substitution). m AF D = m CF E (Vertical Pair Theorem). m C + m AFD = m B + m D (more substitution). m B + m AFD = m B + m D (yet more substitution, since m C = m B). m AFD = m D (algebra). The above angles are opposite the sides AD and AF of triangle F AD. Since the base angles are congruent, by the Isosceles Triangle Theorem, FAD is isosceles with base FD.

5 #8: I ll prove Corollary D. Corollary A is in an assignment. If a line is perpendicular to one of two parallel lines, then it is perpendicular to the other also. Suppose AB and CD are parallel lines, with transversal EF. Suppose EF AB. Then AEF is a right angle (definition of perpendicular). Since the lines are parallel, alternate interior angles are congruent (Property Z (Parallel Postulate)) so DFE = AEF and DFE is a right angle. So EF CD. #10a: Given: L and M are midpoints of AB and AC, D and E are midpoints of AL and AM. Prove: Lengths of the sides of ADE are one fourth that of ABC. Since M is the midpoint of AC, AM = 2 1 AC. Since E is the midpoint of AM, AE = 2 1 AM. So AE = 1 2 AM = 1 2 (1 2 AC) = 1 4 AC (All following from definition of midpoint, plus a little algebra.) Similarly, since L is the midpoint of AB, AL = 1 2 AB. Since D is the midpoint of AL, AD = 1 2 AL. So AD = 1 2 AL = 1 2 (1 2 AB) = 1 4 AB (All following from definition of midpoint, plus a little algebra.) Since L and M are the midpoints of AB and AC respectively, by the Midpoint Connector Theorem, LM = 1 2 BC. Since D and E are the midpoints of AL and AM respectively, by the Midpoint Connector Theorem, DE = 1 2 LM. DE = 1 2 LM = 1 2 (1 2 BC) = 1 4 BC The lengths of the sides of ADE are one fourth that of ABC.

6 #13: Given: PQ = PR = PS, Q P R. Prove: SQR is a right triangle. Since PS = PR, m 1 = m 2 (Isosceles Triangle Theorem). Since PS = PQ, m 3 = m 4 (Isosceles Triangle Theorem). Since Q P R, P is in the interior of Crossbar Theorem). RSQ (it s Thm 3, p almost but not quite the So by the Angle Addition Postulate, m RSQ = m RSP + m m P SQ, i.e. Then (substitution and algebra): SQR is a right triangle. m RSQ=m 1 + m 4 m RSQ = m 2 + m 3 m RSQ + m 1 + m 4 = 180 (Angle Sum Theorem) m RSQ + m RSQ = 180 2m RSQ = 180 m RSQ = 90