Ch 7.1 · Quadrilaterals

Lesson 1 · Quadrilaterals

Page 258

An advertisement's window with one right angle and four unequal sides opens the chapter on four-sided figures, and the first order of business is upgrading a definition. Page 143 tested convexity by extending sides; here is the sharper version: a polygon is convex iff the segment joining any two interior points stays entirely inside. The green segments in the margin pass; the concave dart's red segment escapes.

Two housekeeping facts follow: triangles are always convex (you proved the dent impossible on page 143), and from here on every polygon with more than three sides is assumed convex unless the book says otherwise. That standing assumption quietly powers half the proofs ahead, so note it now.

Lesson 1 · Quadrilaterals

Page 259

Vocabulary, then a theorem with a famous corollary. Sides that meet are consecutive, sides that do not are opposite, and a diagonal joins nonconsecutive vertices. Diagonals from one vertex carve any polygon into triangles, which converts triangle knowledge into polygon knowledge wholesale: for a quadrilateral, one diagonal makes two triangles, so Theorem 24 says the angles total 360°.

The corollary names a shape: equiangular forces four angles of 360/4 = 90 each, and a quadrilateral of right angles is a rectangle (Latin rectangulum, right-angled). Read the iff carefully: the definition demands right angles; the corollary earns that equiangular alone suffices. Definition and characterization are different purchases, and this chapter trades in both constantly.

New here diagonal · rectangle

Worked example · page 260

Questions 1 to 5: proving the 360

ABCD is a quadrilateral. Prove ∠A + ∠B + ∠C + ∠D = 360°, giving each reason.

Question 1: draw diagonal BD; two points determine a line. One segment, two triangles.
Question 2: each triangle's angles total 180 (the Angle Sum Theorem, twice): ∠A + ∠1 + ∠4 = 180 and ∠2 + ∠3 + ∠C = 180.
Question 3: add the equations: the six pieces total 360 (addition property).
Question 4: ∠1 + ∠2 = ∠ABC and ∠3 + ∠4 = ∠CDA, the Betweenness of Rays Theorem reassembling the split corners.
Question 5: substitute the sums back in: ∠A + ∠ABC + ∠C + ∠CDA = 360°. One diagonal, two 180s, and the quadrilateral's budget is fixed forever, the number your kite figures and SAT problems have been spending since page 160.

Worked example · page 261

Questions 26 and 27: the SAT quadrilateral

A quadrilateral has two right angles, and its other two angles measure 2x° and 2y° (each marked as two equal halves x, x and y, y). What can you conclude about 2x + 2y and about x + y?

Theorem 24 taxes every quadrilateral 360°: 90 + 90 + 2x + 2y = 360.
Question 26: subtract the two right angles: 2x + 2y = 180.
Question 27: divide by 2: x + y = 90; the half-angles are complementary.
Note what the figure was hinting with its bisected corners: the two bisecting rays would meet at a right angle, page 121's linear-pair-bisector fact in quadrilateral form. The SAT wanted two subtractions; the geometry offers a bonus theorem to anyone who keeps looking.

Lesson 1 · Exercises, Set II

Page 262

Penrose's two tiles (the kite and the dart, both built on 72-degree angles) tile the plane in infinitely many ways without ever repeating, a discovery that eventually rewrote crystallography. The worked example dissects them. The gemstone questions then launch the generalization machine: diagonals from one vertex of a pentagon, hexagon, and the oval-cut center polygon, counting sides, diagonals, and triangles toward page 263's formulas.

Question 36 deserves its own frame: equilateral does not force equiangular for quadrilaterals (a leaning rhombus refuses), so the triangle corollary from page 159 does not travel. Every polygon family has its own rules; that is why this chapter exists.

Worked example · page 262

Questions 28 to 35: inside the Penrose kite

In the figure, AB = BE = ED = DA and CB = CE = CD, with 72° angles marked at A (twice), D, and E. Which tile is convex, what is ∠BCD, and why are A, C, E collinear?

Question 28: of the two tiles, the kite is convex; the dart's pushed-in corner fails the page 258 segment test (an interior segment can escape through the notch).
Question 29: apply Theorem 24 to the quadrilateral holding the three 72s: ∠BCD = 360 − 72 − 72 − 72 = 144°.
Question 31: with AB = BE = ED = DA and CB = CE = CD, SSS certifies △ADC ≅ △ABC and △EDC ≅ △EBC; the diagonals split the tiles into matched pairs.
Questions 32 to 34: all four triangles are isosceles, and chasing their base angles fills in every corner. At C the angles on one side of the line through A and E total exactly 180, so A, C, E are collinear, the straight-angle test from page 168.
Question 35: the figure folds over line AE (or ACE): line symmetry, with the matched SSS pairs as the two halves. Aperiodic tiling, built from isosceles triangles and one collinearity; five-fold angles hiding in four-sided tiles.

Worked example · page 263

Questions 43 to 51: the n-gon formulas

From the pentagon, hexagon, and oval-cut drawings: how many diagonals from one vertex, how many triangles, what angle sum, and what does each angle of an equiangular n-gon measure?

Question 43: from one vertex you cannot draw diagonals to itself or its two neighbors: n − 3 diagonals. (Check: pentagon 2, hexagon 3, as your drawings show.)
Question 44: those diagonals fan the polygon into n − 2 triangles, one more than the diagonal count.
Question 45, the audit: a quadrilateral gives 4 − 3 = 1 diagonal and 4 − 2 = 2 triangles, exactly page 260's proof.
Questions 46 to 50: each triangle contributes 180, so the angle sum is (n − 2)180: hexagon 720, octagon 1080, n-gon in general. Equiangular shares equally: hexagon 120°, octagon 135° (the stop sign's corner), and in general (n − 2)180/n.
Watch the formula's two limits: n = 3 returns the sacred 180, and huge n pushes each angle toward 180 as the polygon rounds toward a circle. One diagonal fan, and every polygon's angle budget is published in advance.

Worked example · page 264

Questions 56 to 59: the sum that fixes itself

Suppose every quadrilateral's angles total the same number S. A line through X on BC and Y on AD splits ABCD into two quadrilaterals. Derive S = 360.

Write the three assumed equations: the whole quadrilateral gives ∠A + ∠B + ∠C + ∠D = S, and each half quadrilateral gives S too, with the four new angles 1 through 4 at X and Y included.
Question 56: add the two halves' equations: every original angle appears once, plus all four new angles: total 2S (addition).
Question 57: substitute S for the original four angles' sum: S + ∠1 + ∠2 + ∠3 + ∠4 = 2S.
Question 58: subtract S: the four new angles alone total S.
Question 59: but those four angles are two linear pairs along the cutting line at X and at Y: 180 + 180 = 360. So S = 360, forced by consistency alone. The same style of argument, "if a constant exists, it can only be this," runs deep in mathematics, and this is a perfect first specimen.

Lesson 2 · Parallelograms and point symmetry

Page 265

Octopus vision as a lesson opener: shown three shapes, an octopus groups the parallelogram with the concave hexagon and flags the other parallelogram as the stranger, because cephalopod eyes read orientation, not sides. Geometry groups differently: A and B are both parallelograms, quadrilaterals whose opposite sides are parallel, and that definition is the lesson's entire foundation.

The promised cargo: opposite sides equal, opposite angles equal, and a symmetry the octopus would appreciate, point symmetry, the upside-down kind. Page 266 delivers all three. Note that "parallelogram" was constructed on page 227 before it was studied; now the study begins in earnest.

New here parallelogram

Lesson 2 · Theorems 25 and 26

Page 266

A figure has point symmetry when it looks the same turned upside down: rotate 180° about the center and it lands on itself. For points, the definition sharpens: two points are symmetric with respect to a point iff it is the midpoint of their segment. Every parallelogram has this symmetry about its diagonals' crossing, and that is Theorem 26 in disguise: the diagonals bisect each other.

Theorem 25 (opposite sides and angles equal) gets the full proof: one diagonal, alternate interior angles twice, ASA, and a betweenness reassembly for the split angles. Read it beside page 224's proof 40; they are converses meeting in the middle, and together they say parallel opposite sides and equal opposite sides arrive as a package.

New here point symmetry

Worked example · page 267

Questions 10 to 15: the diagonals bisect each other

ABCD is a parallelogram with diagonals AC and BD meeting at E. Supply the reasons that they bisect each other.

Question 10: BC = AD, opposite sides of a parallelogram are equal (Theorem 25, fresh from the previous page).
Question 11: BC ∥ AD, the definition of a parallelogram itself.
Question 12: the parallel sides make two pairs of alternate interior angles across the diagonals: ∠1 = ∠2 and ∠3 = ∠4 (Theorem 19's first corollary).
Question 13: △BEC ≅ △DEA by ASA: two angle pairs flanking the equal sides BC and AD.
Questions 14 and 15: corresponding parts give BE = DE and EC = EA, and equal halves is what bisecting means. E is the midpoint of both diagonals at once: the parallelogram's center of symmetry, certified. Every point-symmetry claim on the next page stands on this proof.

Worked example · page 268

Questions 24 to 30: Newton's parallelogram of forces

Draw AB = 5 cm and AC = 2.5 cm with an 80° angle at A; complete the parallelogram and draw diagonal AD. If force AB is 50 pounds, read off the rest.

Question 24: "Corol." abbreviates corollary, and this one is to Newton's laws of motion: forces add along the parallelogram's diagonal. Even his book design copied Euclid.
Construct with page 225's parallel construction: through C parallel to AB, through B parallel to AC, D at the crossing.
Questions 26 and 27: 5 cm stands for 50 pounds, so 1 cm is 10 pounds, and force AC = 2.5 × 10 = 25 pounds.
Questions 28 and 29: measure the diagonal: AD comes out near 5.9 cm, so the combined force is about 59 pounds. Notice it is less than 50 + 25; forces at an angle partly cancel, which the parallelogram displays and arithmetic alone would hide.
Question 30: the protractor reads roughly 24° between AD and AB: the combined force leans toward the stronger component. Your compass just did vector addition, three centuries before that name existed.

Worked example · page 269

Questions 47 to 50: the hidden equilateral

ABCD is a parallelogram with ∠BAD = 75°; equilateral triangles ABE and BCF are built on two of its sides. Find three congruent triangles and count the equilaterals.

Mark what the figure gives: AE = AB = BE and BF = BC = CF (equilateral sides), plus AB = DC and AD = BC (Theorem 25).
Chase the angles at each equilateral's corner. The parallelogram gives ∠ABC = 105° (supplement of 75). Then ∠EAD = 60 + 75 = 135°, ∠EBF = 360 − 60 − 60 − 105 = 135°, and ∠FCD = 60 + 75 = 135°: three matching 135s.
Questions 48 and 49: △EAD, △EBF, and △FCD match by SAS: each holds one equilateral-triangle side, one parallelogram side, and a 135° between them (EA with AD, EB with BF, FC with CD).
Corresponding parts make ED = EF = FD: triangle DEF is equilateral.
Question 50's count: three equilateral triangles: the two you built and the one that appeared. This is a cousin of Napoleon's theorem, and it works from any parallelogram angle, 75° being merely the book's choice. Symmetry manufactured, not assumed.