Session 05 · The Protractor and Bisection

Lesson 3 · The protractor and angle measure

Page 91

The degree is older than almost everything else you will ever use: the Babylonians chose it more than 4,000 years ago, and no other unit of measurement has lasted remotely as long. The page's two pictures then make an idea you will need all lesson: sweep a ray once around its endpoint and you get a rotation of rays, 360 of the degree marks; a half-turn is a half-rotation, 180, and the familiar semicircular protractor is a picture of exactly that half-rotation.

Watch what the book is setting up. Page 84 turned a ruler into a numbered line; this lesson turns a protractor into a numbered fan of rays, and every idea from Lesson 2 (coordinates, positive differences, betweenness, even the two-column theorem) is about to be replayed with rays in place of points. Predict the parallels now and page 92 will feel like recognition instead of news.

New here rotation of rays

Lesson 3 · The protractor and angle measure

Page 92

The Protractor Postulate (Postulate 4) is the Ruler Postulate said about rays: number the rays of a half-rotation 0 to 180 and positive differences measure angles. That is why the protractor's center must sit on the vertex, and why the 0 mark need not sit on either side: 160 − 60 gives the same 100° that 120 − 20 gives. Every ray gets a coordinate; every pair of rays gets a measure.

Then four definitions you will use daily: an angle is acute below 90°, right at exactly 90°, obtuse between 90 and 180, straight at exactly 180. Notice the boundary cases are knife-edge exact: 90.0001° is already obtuse. The bottom figure starts betweenness of rays, finished on the next page.

New here acute · right · obtuse · straight

Khan Academy Acute, right, and obtuse angles

Worked example · page 93

Questions 4 to 7: rays HE, HI, and HO

Ray HE has coordinate 81, ray HI has 18, ray HO has 180. Which ray is between the other two? Find the three angles, and name the statement that justifies each answer.

Question 4: compare coordinates, not the picture: 18 < 81 < 180, so ray HE is between rays HI and HO. Reason (question 6): the definition of betweenness of rays.
Question 5: subtract coordinates pairwise. ∠IHE = 81 − 18 = 63°.
∠EHO = 180 − 81 = 99°.
∠IHO = 180 − 18 = 162°.
Check the sum: 63 + 99 = 162, and the statement that promised it (question 7) is the Betweenness of Rays Theorem. Also classify while you are here, using page 92: 63° acute, 99° obtuse, 162° obtuse. Drawing the figure with your protractor makes all three subtractions visible.

Lesson 3 · Exercises, Set I

Page 94

A page of angles in the wild. The 1905 schoolbook excerpt (8 to 10) is there to show vocabulary drifting across a century ("half-lines" for rays, "arms" for sides) while the naming rule held steady: vertex letter in the middle, always. The basketball "360" and the one-and-a-half somersault (a 540) make the page 91 point that turning is unlimited even though geometry's angles stop at 180. The board photograph is the sly one: real right angles photograph as acute and obtuse, so measuring a picture is not measuring the thing.

The red sector chart is the page's protractor-postulate moment, and the worked example does it. Note its navigation twist: bearings are numbered clockwise from north, 0 to 360, a circular protractor in working uniform.

Worked example · page 94

Questions 11 to 13: the red sector

A signal light's danger area lies between bearings 045° and 120°. What are those two numbers with respect to the sides of the angle? What is the angle's measure, and what kind of angle is it?

Question 11: the chart is a circular protractor centered on the light, so 45 and 120 are the coordinates of the angle's two sides, exactly as in Postulate 4.
Question 12: measure = positive difference of coordinates: 120 − 45 = 75°.
Question 13: classify with page 92's definitions: 75 is less than 90, so the danger angle is acute.
Self-check the setup: a mariner reads those same numbers off a compass card, and the subtraction they do at sea is the Protractor Postulate verbatim. Runway numbers on page 95 will be the same idea divided by ten.

Worked example · page 95

Questions 26 to 33: the cactus spokes

Rays CS, CP, CK have coordinates 43.2, 57.6, 86.4. Is ray CP between rays CS and CK? Find ∠SCP, ∠PCK, and ∠SCK, write the equation relating them, and name the theorem it illustrates.

Questions 26 and 27: test the definition, not the picture: 43.2 < 57.6 < 86.4, so yes, CP is between CS and CK because its coordinate is between theirs.
Question 28: ∠SCP = 57.6 − 43.2 = 14.4°. Notice 14.4 = 360/25: the cactus carries 25 evenly spaced spokes.
Question 29: ∠PCK = 86.4 − 57.6 = 28.8°, two spoke gaps.
Question 30: ∠SCK = 86.4 − 43.2 = 43.2°.
Questions 31 to 33: 14.4 + 28.8 = 43.2, which is ∠SCP + ∠PCK = ∠SCK, the Betweenness of Rays Theorem: if one ray is between two others, its two angles add to the whole angle. Same skeleton as the light-years on page 89, now in degrees.

Lesson 3 · Exercises, Set II

Page 96

Two runways can share a number only by pointing the same direction, which is what the photograph's 25L and 25R mean: parallel strips, left and right. The NBC peacock divides a half-rotation among feathers (180/9, 180/10, 180/12), and the earth questions turn the clock into a protractor: one rotation in 24 hours is 15° per hour, and the angular minute and second borrow their names from time on purpose.

The pool table is the page's real work. Given the physical fact that a banking ball leaves at the angle it arrived (∠1 = ∠2, and ∠CPX = ∠DPY), questions 41 to 43 prove the two upper angles equal too, and the proof is nothing but this lesson's theorem plus page 79's subtraction property. The worked example walks it.

Worked example · page 96

Questions 41 to 43: the pool-ball proof

Given ∠1 = ∠2 and ∠CPX = ∠DPY, prove ∠3 = ∠4. Justify each of the three listed statements.

Question 41: ∠CPX = ∠1 + ∠3 because ray PA is between rays PX and PC (PX-PA-PC), and likewise ∠DPY = ∠2 + ∠4 because PD-PB-PY. Reason: the Betweenness of Rays Theorem, splitting each big angle at the ball's path.
Question 42: the two big angles are equal (given), so substitute the sums from step 1 for them: ∠1 + ∠3 = ∠2 + ∠4. Reason: substitution.
Question 43: ∠1 = ∠2 is given, so subtract it from both sides: ∠3 = ∠4. Reason: the subtraction property of equality.
Step back and count ingredients: one physical fact about cushions, one theorem about rays, one property of equality. That little proof is the pattern of half the proofs in this book: geometry splits, algebra cancels.

Lesson 3 · Exercises, Set III

Page 97

After question 54 closes the lesson's bookkeeping (you write Theorem 2's other case, mirroring page 89's question 47), Set III serves a legend. This angle puzzle ran in the Washington Post in 1995; the syndicate swore it was solvable, the author forgot how, and three dozen geometers reportedly came up empty. It is known to mathematicians as Langley's problem, and the book asks only that you draw it large, chase the easy angles, and commit to a guess.

Chase them honestly: the triangle angle sum from page 66 hands you several angles almost free, and an accurate drawing will whisper the answer for question 3. The full deductive solution needs one very clever auxiliary line, which is why it beat the professionals; enjoy being stuck where they were.

YouTube MindYourDecisions solves a hardest-easy-problem variant

Worked example · page 97

Questions 1 to 3: cornering the Post puzzle

In the figure, angle A is split 60° and 20°, angle B is split 70° and 10°. Which angles can the triangle angle sum deliver, and what is a good guess for ∠EFB?

Draw it big, as told: the effect of a small figure is the page 12 lesson in reverse; crossings blur and guesses go bad.
Whole triangle first: ∠A = 60 + 20 = 80 and ∠B = 70 + 10 = 80, so the apex angle at C is 180 − 80 − 80 = 20°.
Triangle ABF (the one using all of ∠A and B's lower 70): ∠AFB = 180 − 60 − 80 = 40°. Triangle ABE (all of ∠B and A's lower 60... check which crossing your drawing labels): the same recipe gives ∠AEB = 30°. Each crossing point surrenders to one more angle-sum.
∠EFB refuses the recipe; every triangle containing it has two unknowns. That refusal is the puzzle's fame. Measure your drawing: a good figure reads close to 20°, and 20° is in fact the answer.
Keep the moral, because it is the chapter's: measurement can suggest 20° but cannot prove it, and the proof needs a line nobody asked you to draw. Auxiliary lines arrive officially in a later chapter; this puzzle is why they exist.

Lesson 4 · Bisection

Page 98

A Kentucky man answered a divorce demand for an equal property settlement by sawing the house into "His" and "Hers" halves until a restraining order stopped him. The book's opening photograph is that house, and the lesson is what he thought he was doing: bisection, cutting a thing into two equal parts.

Chapter 1 bisected with straightedge and compass; here the tool is humbler and sharper: fold tracing paper so one endpoint lands on the other, and the crease cannot help but cut the segment at its midpoint. The definition makes it official (a point is the midpoint iff it divides the segment into two equal segments), and notice it is stated in the iff form every definition has carried since page 47.

New here midpoint

Lesson 4 · Bisection

Page 99

Fold one side of an angle onto the other and the crease bisects the angle. But watch the word "equal" in these folding sentences, because the page stops to upgrade it: Euclid's "things which coincide are equal" becomes the word congruent, coinciding exactly when superimposed. Segments are congruent when their lengths are equal, angles when their measures are equal; bisection always produces two congruent parts. This word will eventually carry whole chapters.

The ruler bisects too: a 4.2 cm segment has its midpoint at exactly 2.1 cm, and since the Ruler Postulate promised one point per number, a segment has exactly one midpoint. A statement that falls out of another that directly is a corollary, the page's third new word, and this one is the Corollary to the Ruler Postulate.

New here bisect · congruent · corollary

Worked example · page 100

Questions 12 to 14: the arrow illusion

Points M and N are marked on segment AB between two arrowheads. Which looks like the midpoint? Could both be midpoints? Which two segments along line AB are equal?

Question 12: to nearly every eye, N looks like the midpoint; the inward and outward arrowheads stretch and shrink the halves.
Question 13: both cannot be midpoints, and you can cite the exact statement: the Corollary to the Ruler Postulate says a segment has exactly one midpoint. Two would need one number to own two points.
Now measure with your ruler, as question 14 orders. The measurement says AM = MB: M is the true midpoint, and the segments AM and MB are the equal pair.
Sit with the punchline: your eye voted N, the corollary said one winner only, the ruler crowned M. Perception proposes, deduction disposes; that division of labor is why this chapter keeps postulates between you and your eyes.

Worked example · page 101

Questions 26 to 30: the chiming clock

A clock takes 3 seconds to chime 3 o'clock. The chimes are points A, B, C with coordinates 0, ?, 3, and B is the midpoint of AC. How long to chime 6 o'clock?

Question 26: B is the midpoint of AC, so AB = BC by the definition of midpoint; the segment is cut into two equal parts.
Questions 27 and 28: AC = 3 seconds covers two equal gaps, so each gap is 1.5 seconds and B's coordinate is 1.5. Three chimes make two gaps, not three; that is the puzzle's hinge.
Question 29: six chimes make five gaps of 1.5 seconds each, so the six points sit at A 0, B 1.5, C 3, D 4.5, E 6, F 7.5.
Question 30: the sixth chime lands at 7.5 seconds, so 6 o'clock takes 7.5 seconds, not the "obvious" 6. Gaps, not chimes, carry the time.
Check with the ruler picture: the chimes are points, the gaps are segments, and doubling the chime count does not double the segment count. Counting fence posts versus fence rails, on a clock.

Worked example · page 102

Question 35: B is not the midpoint

In the acetylene molecule, A-B-C with AC > 2AB. Complete the indirect proof that B is not the midpoint of AC.

Beginning assumption, opposite of the claim: suppose B is the midpoint of AC.
If B is the midpoint, then AB = BC. Reason: definition of midpoint.
Because A-B-C, AB + BC = AC (Betweenness of Points Theorem), and substituting AB for BC gives 2AB = AC. Reason: substitution.
The contradiction: 2AB = AC collides with the given fact AC > 2AB. A quantity cannot equal what it exceeds.
Therefore the assumption is false, and B is not the midpoint of AC. Chemistry agrees: the carbon-carbon bond in acetylene really is longer than a carbon-hydrogen bond, and this proof is how you know from lengths alone, without a tape measure small enough for molecules.

Lesson 4 · Exercises, Set II

Page 103

The miter joint proof completes here, and it earns a close read: the carpenter's glued corner becomes the algebraic fact that the bisecting ray sits at exactly half of ∠APB, the same "half of 42 is 21" arithmetic as page 100, now proved rather than measured. The worked example fills its blanks.

The angle-definition discussion (37 to 40) is the book being honest about its own foundations. A straight angle's "vertex" is indistinguishable from any other point on the line, its interior rays fill a whole half-plane, and its bisector is only unique because the Protractor Postulate confines us to one half-rotation; an objector waving the other half-plane's perpendicular ray has a point. Textbook definitions are chosen, not discovered, and the book wants you to see the choosing. The passion flower then sets up page 104.

Worked example · page 103

Question 36: the miter joint

Given PA-PC-PB and that PC bisects ∠APB, complete the proof that ∠1 = ∠APB / 2.

Line 2: ∠1 + ∠2 = ∠APB. Reason: PA-PC-PB and the Betweenness of Rays Theorem; the middle ray splits the whole angle.
Line 4: ∠1 = ∠2. Reason: PC bisects ∠APB, and a bisecting line divides an angle into two equal angles (the page 99 definition).
Line 5: substitute ∠1 for ∠2 in line 2: ∠1 + ∠1 = ∠APB, so 2∠1 = ∠APB. Reason: substitution.
Line 6: divide both sides by 2: ∠1 = ∠APB / 2. Reason: the division property of equality (and 2 ≠ 0, as page 83 taught you to mutter).
Compare this with page 80's questions 11 and 12: same skeleton, then done informally, now done as a listed proof. The chapter has been teaching you one proof in slow motion all along.

Lesson 4 · Exercises, Sets II and III

Page 104

The passion flower's five stamens and three stigmas become a circular protractor problem: equal spacing, collinearity, and two bisectors pin down every ray's coordinate, and the worked example runs the whole inference. Questions 46 and 47 then test whether you will let a pretty picture overrule a definition; neither betweenness nor bisection survives the move to the circular scale unchanged.

Set III's rusty compass (fixed radius, an idea explored by Abul Wefa around 980 A.D.) is the hands-on close: bisect a 5-inch segment with a 1-inch compass. The trick is to build outward from both endpoints symmetrically until arcs finally cross; question 4's answer is the angle whose sides sit 2 compass-widths apart, 60° when the arms are walked out one radius each.

Worked example · page 104

Questions 42 to 45: coordinates for the flower

Ray PA has coordinate 0 on a circular protractor, clockwise. A, P, E are collinear; the five stamens (B, C, E, G, H) are equally spaced; PD bisects ∠EPC and PF bisects ∠GPE. Find all the coordinates.

Question 42: A, P, E collinear puts ray PE half a rotation from ray PA: coordinate 180.
Question 43: five equally spaced stamens divide the full 360 into five gaps of 360/5 = 72°, so ∠EPC = 72°.
Anchor the stamens from E = 180 and step by 72 against the clockwise scale: C = 108, B = 36, and on the other side G = 252, H = 324.
Question 44: bisectors halve the 72s: ∠EPD = ∠FPE = 36°, so D = 144 and F = 216.
Question 45's full list, in scale order: A 0, B 36, C 108, D 144, E 180, F 216, G 252, H 324. Now the two traps: PA "between" PG and PC fails the definition (0 is not between 108 and 252), and PE does bisect ∠GPC, since 252 − 180 = 180 − 108 = 72. One no, one yes, both settled by coordinates rather than by the photograph.