Ch 4.2 · ASA, SAS, and Proof

Lesson 3 · ASA and SAS congruence

Page 146

The Dam Busters raid of 1943: bombs that only worked if released from exactly 60 feet, and no altimeter of the day could deliver 60 feet over dark water. The fix was two flashlights, nose and tail, aimed so their beams crossed exactly 60 feet below the plane. When the crew saw one spot of light on the water, they were at height.

Read why it worked like a geometer: plane body and two beams form a triangle, and fixing two angles (the beam mounts) plus the included side (the plane's length) leaves the triangle no freedom at all. Six parts, but three well-chosen ones determine the rest. That observation, made twice over, becomes the two most important postulates in the book on the next page.

YouTube Restored 1943 footage of the bouncing-bomb trials

Lesson 3 · ASA and SAS congruence

Page 147

First the vocabulary of position: a side is included by the two angles at its ends; an angle is included by the two sides that form it; everything else is opposite. Then the two tracing experiments: copy two angles and their included side, and the triangle finishes itself by extending the angle sides; copy two sides and their included angle, and connecting the ends finishes it.

Those observations become Postulate 5 (ASA) and Postulate 6 (SAS): match two angles and the included side, or two sides and the included angle, and the triangles are congruent. Learn to recite both as full sentences on sight of their abbreviations, as the book insists. Every congruence proof for chapters begins by choosing one of these two doors, and the word "included" is the doorkeeper.

New here included · ASA Postulate · SAS Postulate

Khan Academy Triangle congruence postulates

Worked example · page 148

Questions 5 to 9: the French postulate

"Si deux triangles ont deux côtés égaux chacun à chacun comprenant un angle égal, ils sont égaux." Which of our postulates is this, and what are the French abbreviations likely to be?

Mine the cognates: côtés is "sides" (two of them), angle is angle, comprenant is "including," égaux is "equal." Two sides including an equal angle.
Question 5: sides, included angle, sides: this is SAS, and the marked figure beside it confirms the two ticks and the arc between them.
Question 6, in English, as a full sentence: if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent.
Questions 7 and 8: French abbreviates with its own initials: SAS becomes CAC (côté-angle-côté), and ASA becomes ACA (angle-côté-angle).
Question 9: state ACA in English: if two angles and the included side of one triangle are equal to two angles and the included side of another, the triangles are congruent. Different alphabet, same geometry; Euclid travels better than vocabulary does.

Lesson 3 · Exercises, Sets I and II

Page 149

A 1901 schoolbook figure teaches triangle names by exhibit rather than definition, and questions 16 to 21 make you write the definitions yourself: equilateral (all sides equal), isosceles (at least two sides equal), scalene (no two sides equal), with acute, right, and obtuse recycled from page 92 to describe the angles. Every triangle earns one name from each list, as the figure's six cells show.

The eye-focus triangle is ASA wearing your own face: pupils 3 inches apart, two sighting angles, and the object's position is forced. The worked example draws it, and questions 31 and 32 preview a trade-off (angles at L and R grow as O recedes, ∠O starves toward zero) that rangefinders and astronomers live on.

New here equilateral · isosceles · scalene

Worked example · page 149

Questions 27 to 32: the eye-focus triangle

Draw LR 3 inches long, ∠L = 80° and ∠R = 65°, and extend the sides until they meet at O. How far is O, what is ∠O, and what happens as the object recedes?

Question 27: the drawing is ASA in action: one included side (3 inches), two angles, and the triangle closes itself at O with no further choices. That inevitability is the postulate.
Question 28: measure with your ruler: O lands roughly 4 inches from each eye (your drawing may say 3.9 or 4.2; the point is that it says one thing).
Questions 29 and 30: the protractor reads ∠O ≈ 35°, and the angle sum explains why it had to: 180 − 80 − 65 = 35. Measurement and theorem agreeing is your accuracy check.
Questions 31 and 32: push O away in your head: the sighting angles at L and R open toward 90°, and ∠O shrinks toward 0. Your brain reads that shrinking angle as distance, which is depth perception in one sentence.
Now reread the Dam Busters page: same triangle, flashlights for eyes, and the 60 feet is your 4 inches. One postulate, two instruments, one of them yours.

Lesson 3 · Exercises, Sets II and III

Page 150

Three appliances of the new postulates. The astronomy problem is SAS at planetary scale: two earth radii and the 94° between them determine AB, so your 4-centimeter scale drawing (about 5.9 cm across) reports roughly 5,900 miles, and the equal base angles you measure in question 37 are a fact about isosceles triangles the book will prove soon. The molecule path proves the bounce symmetric with ASA plus a reflexive side.

The pond problem is the worked example: the oldest surveying trick in the book, and the direct ancestor of page 151's cowboys. The match puzzles close the lesson as pure play; "no loose ends" is the page 139 polygon definition enforcing itself on matchsticks.

Worked example · page 150

Questions 43 to 47: measuring the pond

Posts at X, O, A, Y, B satisfy X-O-A, Y-O-B, XO = OA, YO = OB. Why is △XOY ≅ △AOB, and how does that measure the pond?

Question 43: the symbols say O is between X and A, and between Y and B: the two stake lines pass straight through O.
Question 44: betweenness makes ray OA and ray OX opposite rays, likewise OB and OY.
Question 45: two pairs of opposite rays make ∠XOY and ∠AOB vertical angles, equal by Theorem 6 (page 112 still earning).
Question 46: XO = OA and YO = OB are given, and the equal vertical angle sits included between them: SAS, so △XOY ≅ △AOB.
Question 47: corresponding parts are equal, so XY (across the water) equals AB (on dry land), and you measure AB with your feet. Every stake is reachable; the unreachable distance comes out anyway. That is what congruence is for, and the Romans knew it.

Lesson 4 · Congruence proofs

Page 152

One new tool and one model to imitate. The tool: since congruent triangles have all corresponding parts equal by definition, the book licenses the short citation "corresponding parts of congruent triangles are equal," the closing line of nearly every proof from here on. The model: the cowboy proof in nine numbered steps, given to prove, statements to reasons.

Study its architecture the way the last paragraph does: three equation steps (4, 6, 7) feed the ASA step (8), which feeds the conclusion (9); everything else exists to justify those three equations. And note the book's own liberty: several step orders are correct. A proof is a dependency tree, not a ritual sequence, and seeing which steps lean on which is the actual skill.

New here corresponding parts

Worked example · page 153

Questions 1 to 7: the first roof proof

Given: D is the midpoint of AC, and BD ⊥ AC. Prove the rafter angles equal: ∠BAD = ∠BCD.

Questions 1 and 2: mark the figure first, as ordered; marked figures are half the proof. D a midpoint gives AD = DC by the definition of midpoint.
Question 3: BD ⊥ AC makes ∠1 and ∠2 right angles (perpendicular lines form right angles).
Question 4: all right angles are equal (page 118's corollary), so ∠1 = ∠2.
Question 5: BD = BD, the reflexive property: the shared post belongs to both triangles, and citing it is how a proof says so.
Questions 6 and 7: two sides (AD = DC, BD = BD) and the included right angles equal: SAS gives △ABD ≅ △CBD, and corresponding parts make ∠BAD = ∠BCD. Now reread questions 8 to 17 and watch the same truss yield to ASA and then SAS from different givens; that flexibility is the lesson.

Worked example · page 154

Questions 26 to 29: the two forgeries

Example 1 "proves" △ADB ≅ △BDC by ASA; example 2 "proves" AB = AC by SAS. Both figures are accurate. Find each proof's lie.

Question 26: look at the figure: △BDC is visibly longer and thinner than △ADB; no tracing fits. So some step must be false, before you even hunt it.
Question 27: the ASA step needs the equal side included between the two equal angles. In △ADB the side BD is included by ∠ADB and ∠DBA, but the proof used ∠A, and BD is opposite ∠A, not included. Right rule, wrong side: ASA never applied.
Question 28: in example 2's accurate figure, AB is plainly longer than AC; the conclusion is false.
Question 29: step 3 claims ∠DAB = ∠DAC "by reflexive." But those are two different angles; reflexive only says a thing equals itself. The equal-looking citation smuggles in an equality nobody granted, and SAS collapses with it.
The moral, twice over: a proof's citations must fit the figure's actual anatomy: included means included, reflexive means the same part. Checking that fit is what page 51 called auditing the logic, and forgeries are the drill.

Worked example · page 155

Questions 34 to 46: folding the graph paper

△ABC has A(3, 1), B(3, 8), C(7, 1). △DEF has D(3, −1), E(3, −8), F(7, −1); △JKL has J(1, 3), K(8, 3), L(1, 7). Relate them all to △ABC.

Questions 35 and 39: legs by subtraction: AB = 7, AC = 4, and the distance formula gives BC = √(16 + 49) = √65. DEF repeats the same three numbers.
Questions 36 to 38: DEF lives in quadrant IV, each vertex the ABC vertex with its y negated. Folding along the x-axis drops one triangle exactly onto the other: a reflection.
Question 40: right angles at A and at D, equal legs around them: SAS certifies △ABC ≅ △DEF with no tracing paper, only coordinates.
Questions 41 to 43: the y-axis fold lands ABC in quadrant II at G(−3, 1), H(−3, 8), I(−7, 1): x negated this time.
Questions 44 to 46: JKL swaps each vertex's coordinates (x, y) to (y, x), which is the fold along the diagonal line through the origin, the line where x = y. Try it: the crease runs corner to corner through (1,1), (2,2), and the vertices meet. Three folds, three reflections, one congruent triangle each: the plane is full of free congruences, and coordinates name them.

Worked example · page 156

Set III: running the prisoner wheel

Numbers 1, 2, 4, 5, 6, 7, 8, 9 around the circle, 3 at the center of the cut-out disc bearing a diameter and two congruent triangles. Rotate and read four days of groupings.

Day one, disc as drawn: the diameter reads 1-3-2, one triangle picks up 4-5-6, the other 7-8-9. List: 123, 456, 789, matching the guard's sketch.
Rotate until the diameter spans 4 and 9 (with 3 riding the center always): read the new diameter trio and the two new triangle trios, writing them down before turning again.
Repeat with the diameter between 7 and 5, then between 8 and 6. Four readings, four days.
Audit like a geometer: pick any pair (say 2 and 6) and scan your four days; they should share a group exactly once. The congruent triangles enforce this: rotation never changes their spacing, so each turn meets each number pattern once.
The last sentence of the page deserves its exclamation point: this wheel is a baby Latin-square design, and field biologists rotate treatments across plots with exactly this trick. Dudeney built lab equipment and called it a puzzle.