Ch 4.3 · Isosceles and SSS

Lesson 5 · Isosceles and equilateral triangles

Page 157

Euclid opened the Elements by constructing an equilateral triangle and closed it with the five regular polyhedra; Poinsot's great icosahedron of 1809, the spiky star in the photograph, arrived two thousand years later and is stuffed with triangles of every breed. The line drawing beside it is a field guide: △ABC scalene, △DEF isosceles, △GHI equilateral, and the formal iff definitions from page 149 now go on the books.

Mind the fine print you noticed before: isosceles means at least two equal sides, so every equilateral triangle is isosceles, and any theorem proved for isosceles triangles applies to equilateral ones free of charge. That inclusion is about to do real work, because this lesson proves the first theorems in the book that are about triangles themselves.

Lesson 5 · Isosceles and equilateral triangles

Page 158

Angle-side classifications finish (obtuse, right, acute, equiangular), and then comes one of the loveliest proofs in the book. Theorem 9: if two sides of a triangle are equal, the angles opposite them are equal. The trick: flip a tracing of △BCD and it fits back on itself, giving the correspondence △BCD ≅ △CBD, a triangle congruent to its own mirror image, by SAS.

Read the two-column version until the strangeness dissolves: the triangle plays both roles in the congruence, BD = CD serving once as a left side and once as a right. Euclid's own proof of this fact was so hard that medieval students named it the pons asinorum, the bridge of asses, the point where weak students turned back. This flip proof, essentially Pappus's, walks you across the bridge in five lines.

New here equiangular

Worked example · page 159

Questions 1 to 6: the kite figure

AB = AD, and ∠C = ∠BDC. Mark the figure, then hunt equal parts in △ABD and △BCD. Are the two triangles congruent?

Question 1: tick AB and AD, arc ∠C and ∠BDC. Marked figures think for you; unmarked ones lie by omission.
Questions 2 and 3: in △ABD the equal sides AB and AD face ∠ADB and ∠ABD, so Theorem 9 makes ∠ABD = ∠ADB. (Which angle faces AB? The one not touching it: ∠ADB.)
Questions 4 and 5: in △BCD the equal angles ∠C and ∠BDC face sides BD and BC, so Theorem 10 makes BD = BC. One triangle used the theorem, the other its converse; that is the pairing this lesson exists to teach.
Question 6: no. Each triangle is isosceles for its own reason, but nothing matches parts of one to parts of the other: no shared side lengths, no matching angles given. Two isosceles triangles need not be congruent any more than two rich men need be brothers.

Worked example · page 160

Questions 29 to 32: the shuffleboard angles

SH = HU = UF = FL = LE = EB and ∠F = 36°. What kinds of triangles are △SBF, △HEF, △ULF, and find ∠S, ∠B, ∠LUF, ∠ULF, ∠SUL, ∠BLU.

Question 29: each triangle has two equal sides built from the equal segments, so all three are isosceles, nested inside the same 36° point at F.
Question 30: in △SBF the base angles at S and B face the equal sides, and the three angles total 180 (page 66's theorem, still on loan): ∠S = ∠B = (180 − 36)/2 = 72°.
Question 31: △ULF is isosceles with the same 36° apex, so ∠LUF = ∠ULF = 72° too. Same shape, smaller size; the word for that arrives in a later chapter (similarity).
Question 32: ∠SUL sits beside ∠LUF on the straight line SF, a linear pair: 180 − 72 = 108°; likewise ∠BLU = 108°.
Check the arithmetic the cheap way: 72 + 72 + 108 + 108 = 360, the four angles of quadrilateral SULB, which is what any quadrilateral's angles must total (two triangles' worth). The court is self-auditing.

Lesson 5 · Exercises, Set II

Page 161

The corollary proofs (34 and 35) are five-line ladders: equilateral unpacks to two side equalities, Theorem 9 fires twice, substitution splices, and the definition of equiangular closes. Rotten eggs then supply an isosceles triangle you can smell: hydrogen sulfide's two equal bonds give equal base angles of 43.9°, an apex of 92.2°, and question 40's trap: the Pythagorean Theorem speaks only of right triangles, and 92.2 is not 90, close as it looks.

The worked example runs questions 41 and 42, a tiny masterpiece that uses both corollaries and Theorem 9 in four moves. The star puzzle (43 to 45) closes the page; question 44's answer is the page 140 definition again (the star's segments cross, so it is not a polygon).

Worked example · page 161

Questions 41 and 42: the leaning equilateral

△ABD is equiangular and D is the midpoint of AC. △BDC is evidently obtuse. What more can be said about it?

Mark the figure: three arcs in △ABD, tick AD = DC.
Equiangular forces equilateral (the corollary you just proved), so AB = BD = AD.
Chain through the midpoint: BD = AD and AD = DC give BD = DC by substitution. So △BDC is isosceles.
Theorem 9 then makes its base angles equal: ∠DBC = ∠DCB. And the angles have sizes if you want them: ∠BDA = 60 (equiangular), so ∠BDC = 120 (linear pair), leaving 30 each for the base angles.
So the "evident" obtuse triangle is precisely a 120-30-30 isosceles triangle, every claim resting on a named theorem. From one equiangular triangle and a midpoint, the whole figure surrendered; that economy is what the two new theorems buy.

Lesson 5 · Exercises, Sets II and III

Page 162

The pentagram pages reward slow thought: equal star segments do not by themselves force the blue pentagon's sides equal (question 46 wants your why), but add equal star-point angles and Theorem 10 plus congruent corner triangles deliver it (question 47). The drawing exercise builds a triangle inside an equilateral triangle from three 1-inch offsets, and SAS proves the three corner triangles congruent, so △XYZ is equilateral too: symmetry you constructed, then certified.

Escher's lizards close the lesson: three crawling directions, 120° of turn between neighbors, and the blue impostor fails because it is a mirror-image lizard; the tessellation uses rotations only, and page 140 taught you that mirror congruence is congruence of a different handedness. The SAT problem is the worked example.

Worked example · page 162

Question 48: the SAT chain of isosceles triangles

PS is a line segment, PQ = QT = TR = RS, ∠P = ∠S = 40°. Find y, the apex angle ∠QTR.

Left triangle first: PQ = QT makes △PQT isosceles, and the equal sides face equal angles: ∠QTP = ∠P = 40°.
So ∠PQT = 180 − 40 − 40 = 100°, and its linear-pair neighbor ∠TQR = 80°.
Mirror the work on the right: △RST gives ∠RTS = 40°, ∠TRS = 100°, ∠TRQ = 80°.
Middle triangle: ∠TQR + ∠TRQ + y = 180, so y = 180 − 80 − 80 = 20.
Cross-check along the top: the three angles at T (40 + y + 40) must make ∠PTS, and in △PTS the base angles are 40 and 40, so ∠PTS = 100 = 40 + 20 + 40. It closes. Isosceles plus linear pair plus angle sum, three theorems round-robin: that is the entire SAT geometry playbook in one figure.

Lesson 6 · SSS congruence

Page 163

A wooden trestle bridge that made an 1875 traveler hold his breath, and the reason it held anyway: trusses, triangles all the way down. Thread drinking straws into a triangle and it locks; thread a quadrilateral and it slumps into a parallelogram at a touch. Three fixed side lengths leave a triangle no shape choices at all, and four leave a quadrilateral infinitely many.

Do the straw experiment for real if you have never felt it; the lesson's theorem is in your fingers before it is on the page. If the sides determine the shape, then equal sides must force congruence, and unlike ASA and SAS, this one need not be assumed. The book now has enough machinery to prove it, and does, overleaf, with the isosceles theorem you learned five pages ago as the engine.

Lesson 6 · The SSS Theorem

Page 164

Theorem 11, SSS: three sides equal, triangles congruent. The proof is the longest yet and worth a slow read in two sittings. Part one builds a copy: a ray at the right angle (Protractor Postulate), a point at the right distance (Ruler Postulate), and SAS certifies the copy congruent to △DFE. Part two shows the copy congruent to △ABC: the segment BP creates two isosceles triangles, Theorem 9 fires twice, the Betweenness of Rays Theorem adds the angle pieces, and SAS closes. The corollary from page 141 splices the two congruences.

Count what the proof consumed: both measurement postulates, SAS, Theorem 9, betweenness, addition, substitution, the corollary. Every tool this book has built shows up for one job. The theorem itself you will use in one line, as the book promises: say it as a sentence whenever you see SSS.

New here the SSS Theorem

Khan Academy Congruent triangles and SSS

Worked example · page 165

Questions 9 to 14: five equal parts, no congruence

The two accurately drawn triangles have their equal parts marked. Are they congruent? How many pairs of equal parts do they have? Then: do 3, 4, 5, or 6 pairs of equal parts force congruence?

Question 9: no. Look with page 140 eyes: no tracing of one fits the other; the second triangle is visibly larger. Yet the marks are honest.
Question 10: count the marks: three pairs of equal angles and two pairs of equal sides: five pairs of equal parts, without congruence.
How can that be? The equal sides are not in corresponding positions: each equal side sits opposite a different angle in the two triangles. (Triangles like 8, 12, 18 and 12, 18, 27 pull this off: same shape, sides shifted one slot.)
Questions 11 to 13: three pairs can fail (three angles: same shape, any size), four can fail, and five can fail, as the figure in front of you proves.
Question 14: six pairs is the definition of congruence itself, so yes, six cannot fail. The moral, in one sentence: congruence postulates are not about how many parts match but about which, and included-ness is the whole game. ASA, SAS, SSS are the arrangements that work; this figure is the tombstone of the ones that don't.

Worked example · page 166

Questions 15 to 19: why angle bisection works

Bisect a 50° angle with compass and straightedge, labeling the arc crossings A and B and the far crossing D. Draw AD and BD. Prove the construction honest.

Question 16: the construction made CA = CB (one arc from C) and AD = BD (equal arcs from A and B). Equal radii are equal segments; that is all a compass ever says.
Question 17: CD = CD, reflexive; the ray under test belongs to both triangles.
Question 18: three pairs of sides equal (CA = CB, AD = BD, CD = CD): △ACD ≅ △BCD by SSS.
Question 19: corresponding parts make ∠ACD = ∠BCD, and equal halves is what "bisects" means (page 99). The construction is proved, not just trusted.
Now run questions 20 to 25 the same way for the segment: SSS gives equal angles at C, then SAS gives △ACE ≅ △BCE, then AE = BE. Chapter 1's magic tricks are now theorems, which is the whole story of this course so far told in one construction.

Lesson 6 · Exercises, Set II

Page 167

Two linkages, physical proof that SSS is about rigidity. Linkage 1 (two pairs of equal rods) flexes, yet whatever shape it takes, the diagonal AC makes △ADC ≅ △CBA by SSS, so opposite angles stay equal at every setting; what you cannot conclude (the diagonals bisecting each other, the little triangles at E matching) is just as instructive, because SSS has no grip there. The grid exercise catches an imposter: a triangle that looks equilateral computes isosceles (√90, √85, √85), and its twin proves congruent by SSS through arithmetic alone.

Linkage 2 is the worked example: a draftsman's angle-bisector that cannot lie, and the proof needs the corollary from page 100 to finish.

Worked example · page 167

Questions 36 to 43: the bisecting linkage

OA = OB and AD = DB = BC = CA. Whatever shape the linkage takes, prove O, C, D are collinear.

Question 38: OA = OB (given), CA = CB (given), OC = OC: △AOC ≅ △BOC by SSS, at every setting of the hinges. Question 39: so ∠AOC = ∠BOC always.
Question 40: equal halves means line OC bisects ∠AOB, by definition of bisecting.
Question 41: repeat with D: OA = OB, DA = DB, OD = OD: SSS again, so OD also bisects ∠AOB.
Question 42: the Corollary to the Protractor Postulate (page 100) allows an angle exactly one bisecting ray, so OC and OD are the same line.
Question 43: one line through O, C, and D: collinear, at every opening. Squeeze the linkage and the three points glide but never leave their common line; a uniqueness corollary from chapter 3 just became a drawing instrument. Hilbert put this linkage in his book on geometric imagination, and now you know why.

Worked example · page 168

Set III: why the bracing is rigid

Every rod equals the square's side, so every triangle in the figure is equilateral. Fill in the angles, then show X, Y, Z must be collinear using the angles at Y.

Question 1: every triangle of equal rods is equilateral, so every triangle angle in the figure is 60° (equilateral is equiangular, your page 161 corollary), and the square contributes 90s.
At any joint, the angles of the triangles meeting there add up; write 60s and 90s around each lettered joint until the figure is inked. This is pure bookkeeping and worth actually doing.
Question 2: at Y, three angles from the three triangles meeting on the lower side of XZ: 60 + 60 + 60 = 180°.
Angles filling exactly a straight angle force XY and YZ into one line (the straight-angle definition from page 92 running in reverse). So X, Y, Z are collinear not by looks but by arithmetic.
And rigidity follows: X and Z are pinned to the braced corners through chains of SSS-rigid triangles, the collinear strut XYZ cannot hinge, and the square's corners lose their last degree of freedom. Twenty-three rods, all doing geometry; the bridge on page 163 was never in danger.