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Parallel lines
Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, 2001. In particular, see section 4.1, pp 219-223.
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: m6 1 = m6 2. Prove: m6 3 = m6 4.
−→
Since AB and CD are cut by transversal AC with corresponding angles 6 1 ∼
= 6 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
6 3 and 6 4, the corresponding angles must be congruent, and m6 3 = m6 4 (Property F, Parallel
Postulate).
#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, m6 ACB = 31. Find: m6 1, m6 2, m6 3.
31 + 90 + m6 1 = 180, so m6 1 = 59 (Angle Sum Theorem).
Careful for the next bit - yes, it is true that 31 + 90 + m6 2 = 180...but the thing you’re thinking
of is called the linear pair axiom, not the “three angles sharing a line” axiom. Do it right:
m6 ACD + 31 = 180, so m6 ACD = 149 (Linear Pair Axiom).
m6 DCE + m6 ECA = m6 ACD (Angle Addition Postulate).
m6 2 + 90 = 149, so m6 2 = 59.
59 + 90 + m6 3 = 180, so m6 3 = 31 (Angle Sum Theorem).
#3: Given: Figure. Find: x, y, z.
43 + 90 + z = 180, so z = 47 (Angle Sum Theorem).
z + y = m6 ACB (Angle Addition Postulate), so 47 + y = 90 and y = 43.
x + y + 90 = 180, so x + 43 + 90 = 180 and x = 47 (Angle Sum Theorem).
#5: Given: 4ABC equilateral, DE ⊥ BC. Prove: 4ADF is isosceles.
Since 4ABC 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.]
m6 A = m6 B = m6 C
m6 B + m6 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).
m6 C + m6 CF E = 90 (corollary to Angle Sum Theorem).
m6 C + m6 CF E = m6 B + m6 D (substitution).
m6 AF D = m6 CF E (Vertical Pair Theorem).
m6 C + m6 AF D = m6 B + m6 D (more substitution).
m6 B + m6 AF D = m6 B + m6 D (yet more substitution, since m6 C = m6 B).
m6 AF D = m6 D (algebra).
The above angles are opposite the sides AD and AF of triangle 4F AD. Since the base angles
are congruent, by the Isosceles Triangle Theorem, 4F AD is isosceles with base F D.
#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 6 AEF
is a right angle (definition of perpendicular). Since the lines are parallel, alternate interior angles
are congruent (Property Z (Parallel Postulate)) so 6 DF E ∼
= 6 AEF and 6 DF E 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 4ADE are one fourth that of 4ABC.
Since M is the midpoint of AC, AM = 21 AC. Since E is the midpoint of AM, AE = 21 AM. So
1
1 1
1
AE = AM = ( AC) = AC
2
2 2
4
(All following from definition of midpoint, plus a little algebra.)
Similarly, since L is the midpoint of AB, AL = 21 AB. Since D is the midpoint of AL, AD = 21 AL.
So
1
1 1
1
AD = AL = ( AB) = AB
2
2 2
4
(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 = 21 BC.
Since D and E are the midpoints of AL and AM respectively, by the Midpoint Connector Theorem, DE = 12 LM.
1
1 1
1
DE = LM = ( BC) = BC
2
2 2
4
The lengths of the sides of 4ADE are one fourth that of 4ABC.
#13: Given: P Q = P R = P S, Q − P − R. Prove: 4SQR is a right triangle.
Since P S = P R, m6 1 = m6 2 (Isosceles Triangle Theorem).
Since P S = P Q, m6 3 = m6 4 (Isosceles Triangle Theorem).
Since Q − P − R, P is in the interior of 6 RSQ (it’s Thm 3, p 108 - almost but not quite the
Crossbar Theorem).
So by the Angle Addition Postulate, m6 RSQ = m6 RSP + m6 m6 P SQ, i.e.
m6 RSQ = m6 2 + m6 3
Then (substitution and algebra):
m6 RSQ=m6 1 + m6 4
m6 RSQ + m6 1 + m6 4
m6 RSQ + m6 RSQ
2m6 RSQ
m6 RSQ
4SQR is a right triangle.
=
=
=
=
180
180
180
90
(Angle Sum Theorem)