Ch 3.4 · Perpendicular and Parallel

Lesson 7 · Perpendicular and parallel lines

Page 117

Four Corners: the one point in the United States where four state borders cross, and the map is the lesson. Utah, Colorado, Arizona, and New Mexico meet because two lines meet, and those lines are perpendicular: they form a right angle. One right angle is the entire definition; the little square in the figure marks it, and the symbol ⊥ writes it (AB ⊥ CD).

Notice how little the definition demands: one right angle, singular. The claim that all four angles at the crossing are then right angles is real work, owed and paid on the next page, and everything needed for the proof (linear pairs, vertical angles) is last session's inventory. The chapter's machinery is about to close its loop.

New here perpendicular

Lesson 7 · Perpendicular and parallel lines

Page 118

Three results and a definition, at speed. Theorem 7: perpendicular lines form four right angles (one 90 spreads to all four via a linear pair and two vertical pairs). A corollary to the definition of a right angle: all right angles are equal, since all measure 90. Theorem 8 runs the traffic the other way: if a linear pair's angles are equal, the common side is perpendicular to the other two sides' line. Note the proofs are written as paragraphs now, not columns; same skeleton, lighter dress, and you should be able to re-column them.

Then the word that will own later chapters: two lines are parallel iff they lie in one plane and do not intersect, written m ∥ n. Mind both clauses; the same-plane condition is there because space allows lines that neither meet nor run parallel, and page 128's diamond crystal will make that concrete.

New here parallel

Khan Academy Identifying parallel and perpendicular lines

Lesson 7 · Exercises, Set I

Page 119

The refraction figure (sunlight bending as it enters water) is a six-line naming drill, and the worked example runs it, because saying exactly which definition or theorem justifies each line is the chapter's endgame skill. The gymnastics question is sneaky-deep: real parallel bars can fail to be parallel by not being straight, by not being coplanar, or by meeting; the definition has three ways to miss.

The dartboard's ten concurrent lines replay page 114, and the Necker cube closes the page with an illusion that flips while you stare. The cube questions ask which edges "appear" parallel or perpendicular, and that word is honest: on paper, almost none of them are, which is the board photograph of page 94 speaking geometry's oldest caution about pictures.

Worked example · page 119

Questions 1 to 6: name every reason

Light meets water at W. Justify each given statement about the numbered angles.

Question 1: ∠5 + ∠6 = 90° makes them complementary. Reason: the definition of complementary angles, nothing more; when a sum is 90, the word applies.
Question 2: ∠SWR a right angle makes SW ⊥ WR. Reason: the definition of perpendicular (two lines forming a right angle).
Question 3: two right angles are equal. Reason: the corollary to the definition of a right angle, all right angles are equal.
Question 4: two angles supplementary to the same angle (∠RWB) are equal. Reason: Theorem 4. Question 5: a line splitting ∠SWR into equal parts bisects it, the definition of bisecting an angle.
Question 6: ∠AWC = ∠AWE is an equal linear pair, so AB ⊥ CE. Reason: Theorem 8. Six statements, six different citations, and not one measurement; that inventory is what the review pages will assume you own.

Lesson 7 · Exercises, Sets I and II

Page 120

Drying mud cracks at right angles, and the worked example proves it from one physical tendency. Then Set II's "deceptive figure" earns a slow look: ∠1 is right but its neighbors are not, which would contradict three theorems, except that nothing in the figure is actually a pair of straight lines through O. Every theorem's hypothesis names its price (linear pair, vertical angles, perpendicular lines), and a figure that never pays is owed nothing. That is question 23 through 27 in one sentence.

The linear-pair bisection construction at the bottom continues onto page 121, where it turns into the session's prettiest result; do the construction now with care and tomorrow's questions answer themselves.

Worked example · page 120

Questions 17 to 22: why mud cracks meet square

Cracks AB and CD meet at B, making ∠1 and ∠2. What are the two angles with respect to sides and measures, and what does their tendency to be equal prove about the cracks?

Questions 17 and 18: the two angles share the side along crack BD, and their other sides are opposite rays along AB. By sides they are a linear pair; by measures, Theorem 5 makes them supplementary.
Question 19: mud tends to crack so the pair is equal. Equal and summing to 180 forces each to 90°.
Question 20: a 90° angle is a right angle, by definition.
Question 21: one right angle between the cracks makes AB ⊥ CD, the definition of perpendicular.
Question 22: the theorem holding the whole chain is Theorem 8, if the angles in a linear pair are equal then their sides are perpendicular. Drying mud minimizes stress, stress balances when the pair equalizes, and Theorem 8 converts that physics into the right angles the photograph shows. Soap films argued the same way on page 95.

Worked example · page 121

Questions 32 to 38: bisectors of a linear pair

OX bisects ∠AOB and OY bisects ∠BOC, where ∠AOB and ∠BOC are a linear pair; the half-angles are marked m, m, n, n. Show ∠XOY is a right angle.

Question 32: ∠AOB and ∠BOC are a linear pair, so they are supplementary by Theorem 5.
Question 33: the whole straight angle is 2m + 2n, and supplementary means 2m + 2n = 180°.
Question 34: divide both sides by 2: m + n = 90°.
Questions 35 and 36: ∠XOY is exactly one m plus one n, so ∠XOY = 90°, a right angle. Questions 37 and 38: one right angle between them makes OX ⊥ OY, by the definition of perpendicular.
Say the result in words and enjoy it: however lopsided the linear pair (10 and 170, 89 and 91), its two bisectors meet at exactly 90°. Check it on your page 120 construction with a protractor, then notice the proof needed no protractor at all.

Worked example · page 122

Set III: how to attack the FD puzzle

Arrange the five pieces to form the letter F. A strategy walk, without the final arrangement.

Inventory the target first: a block letter F is all right angles, roughly ten of them, with every edge parallel or perpendicular to every other. That is the lesson's vocabulary as a shopping list.
Inventory the pieces: count each piece's square corners and its slanted edges. Slanted edges cannot lie on the F's boundary, so every slanted edge must be buried against another piece's matching slant.
Pair the slants by length, exactly as Dudeney's pieces paired on page 23. The long slant needs a long partner; that constraint alone fixes most of the layout.
Place the largest piece to carry the F's tall stem, then let the paired slants dictate the two arms. Expect several dead ends; the 1933 box bragged about the fifty minutes.
When it closes, check your solution like a proof: every boundary edge parallel or perpendicular to the stem, no slant showing. Then question the vocabulary: how many pairs of parallel edges does your finished F contain?

Chapter review · Summary

Page 123

The Charles Addams cartoon (a skier's tracks passing on both sides of the tree) is the perfect review mascot: your eye says impossible, and this chapter taught you to ask for the proof either way. Beneath it, the whole chapter in one column: twenty Basic Ideas, two postulates, eight theorems, three corollaries. Self-test the way you did on page 71: say each item's content aloud before trusting that you know it, and chase the page number of any that hesitate.

See the architecture before diving into exercises: two postulates (Ruler, Protractor) prop up two betweenness theorems, which prop up the bisection corollaries, the angle-pair theorems, and finally perpendicularity. Chapter 2's tower diagram on page 61 is no longer a diagram; you have now lived in it for a chapter.

Worked example · page 124

Questions 10 to 14: the old quadrant

Rays CE and CD have coordinates 65 and 18 on the quadrant's scale. Find ∠ECD; write "CS is between CE and CD" in symbols; if CS bisects ∠ECD, find ∠ECS, the coordinate of CS, and check with ∠SCD.

Question 10: positive difference of coordinates: ∠ECD = 65 − 18 = 47°.
Question 11: CE-CS-CD, the betweenness-of-rays notation from page 93.
Question 12: a bisector halves the angle: ∠ECS = 47/2 = 23.5°.
Question 13: step down from CE's coordinate: 65 − 23.5 = 41.5.
Question 14: check the other half: ∠SCD = 41.5 − 18 = 23.5°, equal halves as a bisector demands. Instrument, postulate, corollary, all in one woodcut; the mathematics aged better than the astronomy.

Worked example · page 125

Questions 17 to 20: the SAT ray figure

∠AOC = 70°, ∠BOD = 80°, ∠AOD = 110°. Put coordinates on the rays and find ∠BOC.

Question 17: install the Protractor Postulate: OA gets 0, and since ∠AOC = 70, OC gets 70.
Question 18: ∠AOD = 110 puts OD at coordinate 110.
Question 19: OB must sit ∠BOD = 80 away from OD: 110 − 80 = 30. (The figure shows B between A and C, so 30 is the sensible root.)
Question 20: ∠BOC = 70 − 30 = 40°.
Check the whole line of reasoning by adding parts: ∠AOB + ∠BOC + ∠COD = 30 + 40 + 40 = 110 = ∠AOD. Coordinates turned four overlapping angles into subtraction, which is why the SAT liked this figure and why the Protractor Postulate earns its keep under time pressure.

Worked example · page 126

Questions 34 to 36: the stilt walker

A 6-foot person on 6-foot stilts stands on cleats x feet up: S-F-T-H marks stilt bottom, feet, stilt top, head, with ST = FH = 6. How tall is the arrangement, what is TH, and complete the proof that SF = TH.

Question 34: the head stands the person's height above the feet: x + 6 feet up.
Question 35: from stilt top (6) to head (x + 6) is x feet; curious and true: the head clears the stilts by exactly the cleat height.
Question 36, the proof: S-F-T and F-T-H are given; the Betweenness of Points Theorem turns them into SF + FT = ST and FT + TH = FH.
ST = FH is given (both are 6 feet), so substitution gives SF + FT = FT + TH.
Subtract the shared FT: SF = TH, cleat height equals head clearance, the same subtraction that measured the extension ladder on page 89 and the acetylene bond on page 102. One argument, three costumes, full marks if you can now write it cold.

Chapter review · Exercises, Set II

Page 127

The chair proof (39 and 40) is the review's best indirect argument: assume the back perpendicular to the seat, chase the algebra to y = x, and collide with the design rules (x between 5 and 8, y between 20 and 25 can never be equal). Conclusion: a comfortable chair's back is never square to its seat, proved the same way page 89 un-lined the stars.

Graphite's flat hexagons then take the stage: three equal angles around each atom force 120° each, and the proof in question 45 has the angles of the hexagon summing to 720°. Soft graphite is flat sheets sliding; page 128's diamond will be the same carbon refusing to stay in a plane. The little SAT problem in the middle is the worked example, because its entire difficulty is drawing the right figure.

Worked example · page 127

Questions 41 and 42: four points on a line

A, B, C, D are on a line; D is the midpoint of BC. AB = 10, AC = 2, BC = 12. Draw the figure and find AD.

Do not trust the alphabet to give the order; let the numbers place the points. AB = 10 and AC = 2 with BC = 12 means B and C sit on opposite sides of A, since 10 + 2 = 12.
Install coordinates, Ruler Postulate style: A = 0, C = −2, B = 10. Check: BC = 10 − (−2) = 12.
D is the midpoint of BC, so its coordinate is halfway from −2 to 10: (−2 + 10)/2 = 4.
Question 42: AD = 4 − 0 = 4.
The trap this SAT problem sold: sketching A-B-C-D in alphabetical order makes AD unanswerable. The definition of betweenness, not the lettering, dictates the picture; that was page 85's first lesson, and it is the review's last word on coordinates.

Worked example · page 128

Questions 50 to 57: quick time and double time

Quick time: 120 steps per minute, each 30 inches. Find the nth step's coordinate and the marcher's speed in steps per minute, inches per minute, and miles per hour; then compare double time (180 steps of 36 inches).

Questions 50 to 52: each step adds 30 inches, so the fifth step ends at 150, the 100th at 3,000, the nth at 30n; the seven-league boots formula from page 87, in inches.
Questions 53 and 54: 120 steps per minute is given, and 120 × 30 = 3,600 inches per minute.
Question 55: convert in stages: 3,600 inches is 300 feet, so 300 feet per minute; times 60 is 18,000 feet per hour; divide by 5,280 to get about 3.4 miles per hour.
Question 56: double time gives 180 × 36 = 6,480 inches per minute.
Question 57: 6,480 ÷ 3,600 = 1.8, so "double time" is only 1.8 times quick time; the name doubles the step count while the step grows just a fifth. Marmaduke Multiply, waiting on page 130, would approve of checking the arithmetic behind a name.

Algebra review · Operations with polynomials

Page 129

Chapter 3's algebra review moves up a weight class from chapter 2's: polynomials. Adding and subtracting is collecting like terms; multiplying is every-term-times-every-term, which is the distributive rule applied twice, as the color-coded first example shows in slow motion and the "briefer solution" shows at full speed.

You have already met these objects wearing geometry: the binomial square (a + b)² was a literal square on page 81, and the quadratic formula's proof on page 82 ran on exactly these moves. The book reviews them here because the chapters ahead will write lengths and angles as expressions like these and expect the multiplication to be reflex. Watch the sign errors in subtraction, Example 4's little minefield.

Worked example · page 130

Question 20: a three-factor product

Multiply (x + 2)(x − 3)(x + 6).

Two at a time, always. Start with the first pair: (x + 2)(x − 3) = x² − 3x + 2x − 6 = x² − x − 6.
Now multiply the result by (x + 6), every term by every term: (x² − x − 6)(x + 6) = x³ + 6x² − x² − 6x − 6x − 36.
Collect like terms: x³ + 5x² − 12x − 36.
Check with a cheap substitution: at x = 1 the factors give (3)(−2)(7) = −42, and the answer gives 1 + 5 − 12 − 36 = −42. A one-number check catches most sign slips, and it is the page 81 "check by substitution" habit applied to yourself.
The same two-stage discipline handles question 23's cube: square first, then multiply by the third factor. No shortcut survives contact with (x + 4)³ on the first try; the written middle terms do.