Geometry

Flowchart proofs and choosing a format

The same theorem can be proved in a table, in sentences, in boxes joined by arrows, or with coordinates. How do the arrows show which step leans on which, which form should you write, and does the choice change what has been proved?

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Flowchart proofs and choosing a format

Why this exists

A proof is an argument: a chain of statements, each with a reason pointing back to things already settled. Geometry writes it down in four common ways: the two-column proof, the paragraph proof, the flowchart proof and the coordinate proof.

Each form shows some things and hides others. A two-column proof writes its statements in one order, top to bottom, but most steps use only one or two of the lines above them, and two steps that have nothing to do with each other sit one above the other as if the second came out of the first. A paragraph can bury a reason inside a sentence. And one route can need facts the reader has not met when another needs none. Knowing all four, and how to carry a proof between them, lets you pick the form that shows the argument best.

Definition

A two-column proof numbers the statements down the left and writes a reason beside each, naming the earlier lines it uses. Giving a reason for every step is the skill under it.
A paragraph proof writes the same statements and reasons as sentences. The reason rides on a word such as “by”, “because” or “so” (writing a proof as prose).
A flowchart proof, or flow proof, puts each statement in a box with its reason written under it, and draws an arrow into a box from every statement its reason uses. The Given is at the top, the Prove at the bottom, and following the arrows from one to the other is following the argument.
A coordinate proof places the figure on the coordinate plane, gives letters to the coordinates that are free to vary, and proves the theorem by computing (coordinate proofs).

The first three are synthetic proofs: three layouts of one argument from definitions, postulates and theorems about the figure itself. A coordinate proof is a different argument, which turns points into numbers using the distance formula, the midpoint formula and slope.

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)

Flowchart proof: one proof drawn two ways

Take the theorem that vertical angles are congruent, set up as in Given and Prove.

Given. Lines l and m cross, forming \angle 1, \angle 2, \angle 3 and \angle 4.

Prove. \angle 1 \cong \angle 3.

Figure 1Lines l and m cross. Nothing is marked, since the Given says only that they cross.

Plan. \angle 1 and \angle 3 each form a linear pair with \angle 2, so each is what \angle 2 leaves over from 180^\circ.

As a two-column proof:

Statements	Reasons
1. l and m cross	given
2. m\angle 1 + m\angle 2 = 180^\circ	Linear Pair Theorem: line 1
3. m\angle 2 + m\angle 3 = 180^\circ	Linear Pair Theorem: line 1
4. 180^\circ = m\angle 2 + m\angle 3	symmetric property of equality: line 3
5. m\angle 1 + m\angle 2 = m\angle 2 + m\angle 3	substitution: line 4 into line 2
6. m\angle 1 = m\angle 3	subtraction property of equality: line 5
7. \angle 1 \cong \angle 3	definition of congruent angles: line 6

Line 5 puts m\angle 2 + m\angle 3 in place of 180^\circ. Substitution lets the right side of an equation replace its left side, so line 4 first turns line 3 around.

The same proof as a flowchart:

Figure 2The same seven statements as the table. The arrows show that step 5 uses steps 2 and 4, and that steps 2 and 3 do not use each other.

The statements and reasons are the same seven. What the flowchart changes is how the links are shown. The table records them as line numbers after each reason; the flowchart draws each one as an arrow, so the reason under a box no longer needs the numbers. Step 5 has two arrows into it, from steps 2 and 4, because substitution needs both equations. Steps 2 and 3 sit side by side with no arrow between them, so neither uses the other, and the table could list them in either order, which nothing in the table itself makes plain. The table has its own strengths: it is compact, it reads top to bottom like sentences, and each line can be checked where it stands. Proving vertical angles congruent builds this theorem more carefully from the definitions.

Intuition

Strip a synthetic proof down to what matters: the statements, and for each one the statements its reason uses. A two-column proof writes those links as line numbers and picks one order. A flowchart draws the links as arrows and picks no order. A paragraph picks an order and turns the links into words. Moving between them is copying: nothing is added or lost.

That is why a flowchart is the form to reach for when a proof has branches, facts found separately and then put together; while planning, because a box can be put down anywhere and joined up later; and to check a proof, because a box that leads nowhere, or a step with nothing feeding it, shows at once.

A coordinate proof keeps only the Given and the Prove. The steps between are new, and the hard part moves to the start: where the figure is put.

Results

Building a flowchart proof. A few moves, each with a reason.

Put the Given at the top and the Prove at the bottom. A given needs no earlier statement, so nothing points into it; its reason is “given”, and here it is shaded. The Prove is where every path ends, and here it has a double border.
Write one statement per box and its reason under it. The reason is a definition, postulate, theorem or property already accepted, exactly as in a two-column proof.
Draw an arrow into a box from each statement its reason uses, and from no other. The reason decides the arrows. Substitution needs two equations, so two arrows come in; the definition of congruent angles turns one equation into one congruence, so one arrow does. A box with no arrow in must be a given or a statement whose reason needs no earlier statement: the reflexive property, as in \overline{CD} \cong \overline{CD}, or a theorem whose if-part is a position fact read from the figure, as what a diagram may assume allows.
Work from both ends. Forwards: what does the Given let you write? Backwards: what would let you write the Prove? A box can go down before you know what will join it, and the gap closes from both sides.
Read the finished chart. Every box needs a path of arrows down to the Prove; a box without one is true but unused, and belongs in the scratch work, not the proof. No arrows may run in a loop. And any order that writes each box after every box pointing into it is a correct two-column proof of the same theorem.

Carrying a synthetic proof from one form to another.

From	To	How
two columns	paragraph	each row becomes a clause, with its reason after “by” or “because”
two columns	flowchart	each row becomes a box, with an arrow in from each line its reason cites
flowchart	two columns	write the boxes in any order that puts each one below every box pointing into it
paragraph	two columns	one row per claim; “so”, “since” and “in the same way” mark where a reason must be found

When each form is clearest.

Two columns: clearest when every line must be checked on its own, as in learning and grading; weakest when the proof is long and its plan is lost among the rows.
Paragraph: clearest when the idea is short, or the proof assumes something and later drops it; weakest when a reader has to find a reason the sentence left out.
Flowchart: clearest when separate facts are found and then joined, and for planning; weakest when the proof is long and its arrows cross.
Coordinates: clearest for lengths, midpoints, and parallel or perpendicular sides; weakest for angle measures, or when no placement keeps the figure general.

A coordinate proof proves the theorem only if its placement is general: every figure that fits the Given can be moved to sit on the axes that way.

Why it is true

The three synthetic forms stand or fall together. A proof is correct when each reason fits its statement and uses only statements already established. Neither condition mentions layout. “Already established” means “on a line above” in a table and “joined by a path of arrows” in a flowchart; any order that respects the arrows is a correct table, and a paragraph is one such order in sentences.

A general placement proves the theorem for every figure. Any figure can be slid, turned and, if needed, flipped so that a chosen vertex sits at the origin and a chosen side lies along the positive x-axis. Those moves change no length, angle or midpoint, and keep parallel lines parallel (rigid motions). Give every coordinate still free a letter, with no condition the Given does not force, and one proof covers every figure at once, as \triangle ABC does in a Given and Prove.

Worked examples

Example 1

Two branches and one join

Write a flowchart proof that a ray inside a right angle splits it into two complementary angles.

Solution

Given. \angle ABC is a right angle; D is inside \angle ABC.

Prove. \angle ABD and \angle DBC are complementary.

Figure 3The square records the first given; D drawn inside the angle records the second.

Plan. The two parts add up to the whole angle, and the whole angle is 90^\circ.

Figure 4One branch from each given, joined by substitution.

Each given starts a branch. The definition of a right angle turns the first into m\angle ABC = 90^\circ. The angle addition postulate gives m\angle ABD + m\angle DBC = m\angle ABC, and it needs the second given, because the postulate applies only to a ray inside the angle. Neither of these two boxes uses the other, so they sit side by side, and a two-column proof could write them in either order.

The branches join where substitution puts 90^\circ in place of m\angle ABC. Two angles whose measures add to 90^\circ are complementary by definition, which is the Prove.

Example 2

A box with nothing above it

Lines l \parallel m are cut by a transversal t, with the angles numbered as in the figure. Using the Corresponding Angles Theorem, prove that the alternate interior angles \angle 3 and \angle 6 are congruent.

Solution

Given. l \parallel m; t crosses l and m, with the angles numbered as in the figure.

Prove. \angle 3 \cong \angle 6.

Figure 5The arrowheads mark l \parallel m. No two angles are marked equal, since that is what proofs about these lines reach.

Plan. Go through \angle 2. It matches \angle 6 by position, and \angle 3 by sitting opposite it.

Figure 6The vertical angles box needs nothing above it, so no arrow runs into it.

Only one box uses the parallel lines. Corresponding angles \angle 2 and \angle 6 are congruent because l \parallel m, so an arrow runs to that box from the Given. The box \angle 3 \cong \angle 2 has no arrow in. The Vertical Angles Theorem needs only that \angle 3 and \angle 2 sit opposite each other where t crosses l, which is where the set-up puts them, and it would hold whether or not the lines were parallel. The chart shows that at a glance: the long arrow on the right carries a fact that never touched l \parallel m.

The two congruences share \angle 2, so the transitive property joins them. On this map the order runs one way only: the parallel postulate, then corresponding angles, then alternate interior angles, so this proof may cite corresponding angles, never the reverse. The two-column version is proving alternate interior angles congruent.

Example 3

Building from the bottom up

Prove the exterior angle theorem with a flowchart, starting from the Prove.

Solution

Given. \triangle ABC, with \overline{BC} extended past C to D; \angle 1, \angle 2 and \angle 3 are the triangle’s angles at A, B and C, and \angle 4 is \angle ACD.

Prove. m\angle 1 + m\angle 2 = m\angle 4.

Figure 7\angle 4 is the exterior angle at C. No measures are marked.

Plan. Both sides of the Prove are what \angle 3 leaves over from 180^\circ.

Put the Prove in a box at the bottom and ask what could come straight before it. An equation with m\angle 3 added to both sides would do, by the subtraction property:

m\angle 1 + m\angle 2 + m\angle 3 = m\angle 3 + m\angle 4

That box goes above the Prove. What gives it? Two expressions equal to the same number. The left side is the sum in the Triangle Angle Sum Theorem, 180^\circ. The right side is a linear pair, since B, C and D lie on one line, and that is 180^\circ too. Substitution then puts m\angle 3 + m\angle 4 in place of 180^\circ in the triangle’s equation, and for that the linear pair equation must have 180^\circ on its left: 180^\circ = m\angle 3 + m\angle 4, by the symmetric property, is one more box. The two theorem boxes need only the Given, so the Given goes on top with an arrow to each, and the gap has closed.

Figure 8Built from the bottom box upwards, it ends up the same shape as the vertical angles chart.

Working down from the Given alone, it is easy to write true statements that lead nowhere, such as m\angle 2 + m\angle 3 < 180^\circ. Working up from the Prove, every box is there because the box below it asked for it. The two-column version, and what the theorem is good for, is proving the exterior angle theorem.

Example 4

A statement used twice

Prove that if two angles form a linear pair and are congruent, both are right angles.

Solution

Given. \angle 1 and \angle 2 form a linear pair; \angle 1 \cong \angle 2.

Prove. \angle 1 and \angle 2 are right angles.

Figure 9The matching arcs record \angle 1 \cong \angle 2. The square is left off, because right angles are the Prove.

Plan. The two measures add to 180^\circ and are equal, so each is 90^\circ.

Figure 10m\angle 2 = m\angle 1 is used twice, once near the start and once near the end.

Each given starts its own branch. The linear pair gives m\angle 1 + m\angle 2 = 180^\circ, and the congruence gives m\angle 1 = m\angle 2. The plan is to put m\angle 1 in place of m\angle 2, and substitution lets the right side of an equation replace its left side, so the symmetric property first turns the equation around: m\angle 2 = m\angle 1. Substituting then gives m\angle 1 + m\angle 1 = 180^\circ, which needs both equations, so two arrows come in. Combining like terms is its own step, simplify, giving 2\,m\angle 1 = 180^\circ, and dividing by 2 gives m\angle 1 = 90^\circ.

The Prove names both angles, so \angle 2 needs its 90^\circ too. That comes from m\angle 2 = m\angle 1 again, now substituting 90^\circ for m\angle 1. So that box has two arrows leaving it. In a two-column proof it would sit several lines above its second use, with nothing pointing back to it.

Example 5

How many two-column proofs?

How many different two-column proofs does the flowchart of Example 4 stand for? A statement may be written once everything its reason uses is above it.

Solution

m\angle 1 + m\angle 1 = 180^\circ needs the linear pair equation and m\angle 2 = m\angle 1, and those need the boxes above them, so the first five lines are the two givens, their two equations and the symmetric box, in some order. After that the order is forced: m\angle 1 + m\angle 1 = 180^\circ, 2\,m\angle 1 = 180^\circ, then m\angle 1 = 90^\circ, then m\angle 2 = 90^\circ, which uses it, then the last box, which uses both.

So count the orders of the first five. Call the linear pair given G_1 and its equation A, the congruence given G_2, its equation B and the symmetric box S. G_1 must come before A, and G_2 before B before S, and nothing else is fixed. Choose which 2 of the 5 places hold G_1 and A, in that order; the other three hold G_2, B and S, in that order. That gives

\binom{5}{2} = 10

orders: G_1 A G_2 B S, G_1 G_2 A B S, G_1 G_2 B A S, G_1 G_2 B S A, G_2 G_1 A B S, G_2 G_1 B A S, G_2 G_1 B S A, G_2 B G_1 A S, G_2 B G_1 S A and G_2 B S G_1 A. One flowchart, ten equally correct tables. If a course wants every given first, G_1 G_2 or G_2 G_1 must be followed by A B S, B A S or B S A, which leaves 6.

Example 6

The diagonals of a rectangle, by coordinates

Prove that the diagonals of a rectangle are congruent. Write the proof as a paragraph.

Solution

Given. ABCD is a rectangle.

Prove. AC = BD.

Figure 11One corner at the origin and two sides along the axes. The letters a and b can be any positive lengths.

Plan. Put the rectangle on the axes, find both diagonals with the distance formula, and compare.

Place A at the origin with \overline{AB} along the positive x-axis, so B = (a, 0) where a = AB. The angle at A is right, so \overline{AD} lies along the y-axis, upward after a flip if needed: D = (0, b) where b = AD. The right angles at B and D make \overline{BC} vertical and \overline{DC} horizontal, so C has x-coordinate a and y-coordinate b: C = (a, b). By the distance formula,

AC = \sqrt{(a - 0)^2 + (b - 0)^2} = \sqrt{a^2 + b^2}

BD = \sqrt{(0 - a)^2 + (b - 0)^2} = \sqrt{a^2 + b^2}

Both diagonals equal \displaystyle \sqrt{a^2 + b^2}, so AC = BD.

The letters a and b stand for any two positive lengths, so the proof covers every rectangle; for one measuring 6 by 8, each diagonal is \displaystyle \sqrt{36 + 64} = 10. Each step is one short computation, which is why a paragraph suits it.

Example 7

The same proof in two columns and as a flowchart

Rewrite Example 6 as a two-column proof and as a flowchart. What does each show that the paragraph does not?

Solution

Each sentence that makes a claim becomes a row, and the words that carried its reason become the reason.

Statements	Reasons
1. ABCD is a rectangle	given
2. A(0, 0), B(a, 0), C(a, b), D(0, b)	placed on the axes: line 1, right angles at A, B and D
3. \displaystyle AC = \sqrt{a^2 + b^2}	distance formula: line 2
4. \displaystyle BD = \sqrt{a^2 + b^2}	distance formula: line 2
5. \displaystyle \sqrt{a^2 + b^2} = BD	symmetric property of equality: line 4
6. AC = BD	substitution: line 5 into line 3

Figure 12The two diagonals are found separately and meet only in the last box.

The table gives the placement a row of its own, so its reason is checked like any other; the paragraph argued it mid-sentence. It also shows a step the paragraph’s “so” skipped: row 6 puts BD in place of \displaystyle \sqrt{a^2 + b^2}, which substitution allows only from row 5, row 4 turned around. The flowchart shows the shape: the two diagonals are found in boxes with no arrow between them, and only the last box uses both. So the chart stands for 3 different two-column proofs: after the given and the placement, the AC box can go before, between or after the BD box and its turned-around form, and AC = BD comes last.

Example 8

The same theorem without coordinates

Sketch a synthetic proof that the diagonals of a rectangle are congruent, and compare it with Example 6.

Solution

\triangle ABC and \triangle DCB share \overline{BC}, and their angles at B and C are right, since a rectangle has four right angles (properties of a rectangle). If also AB = DC, two sides and the angle between them match, the triangles are congruent, and matching sides give AC = DB (proving parts congruent after the triangles). As a flowchart it is three branches, one per pair of parts, joining at the congruence.

Two of those steps rest on facts that come later on this map, so neither is a reason yet: that opposite sides of a parallelogram, and so of a rectangle, are congruent (proving opposite sides congruent), and the SAS postulate (the congruence criteria). Until then the sketch is a plan. Example 6 needed only the right angles and the distance formula, so it is the proof that can be written now. The synthetic one says why: the diagonals are matching sides of two triangles that fit exactly on each other.

Example 9

The midsegment, by coordinates

Prove that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. Write it in two columns, then draw the length half as a flowchart.

Solution

Given. \triangle ABC; D is the midpoint of \overline{AB}; E is the midpoint of \overline{AC}.

Prove. \overline{DE} \parallel \overline{BC} and DE = \tfrac{1}{2}BC.

Figure 13Factors of 2 in the placement keep the midpoints free of fractions.

Plan. Put B at the origin and C on the positive x-axis, flipping if needed so A is above it. The midpoint formula halves coordinates, so give them a factor of 2: C = (2a, 0) and A = (2b, 2c), with a > 0, c > 0 and b any number. These are the three free choices left: the length of \overline{BC} and the two coordinates of A.

Statements	Reasons
1. D is the midpoint of \overline{AB}	given
2. E is the midpoint of \overline{AC}	given
3. B(0, 0), C(2a, 0), A(2b, 2c), with a > 0 and c > 0	placed on the axes
4. D(b, c) and E(a + b, c)	midpoint formula: lines 1, 2 and 3
5. DE = a	distance formula: line 4
6. BC = 2a	distance formula: line 3
7. \tfrac{1}{2}BC = a	division property of equality: line 6
8. a = \tfrac{1}{2}BC	symmetric property of equality: line 7
9. DE = \tfrac{1}{2}BC	substitution: line 8 into line 5
10. the slopes of \overline{DE} and \overline{BC} are both 0	slope formula: lines 3 and 4
11. \overline{DE} \parallel \overline{BC}	equal slopes: lines 3 and 10, since c > 0 keeps \overline{DE} off the x-axis

Row 9 puts \tfrac{1}{2}BC in place of a, so row 8 first turns row 7 around to put a on the left. The computations behind rows 4, 5 and 10:

D = \left(\frac{0 + 2b}{2},\ \frac{0 + 2c}{2}\right) = (b, c)

E = \left(\frac{2b + 2a}{2},\ \frac{2c + 0}{2}\right) = (a + b, c)

DE = \sqrt{(a + b - b)^2 + (c - c)^2} = \sqrt{a^2} = a

The slope of \overline{DE} is \displaystyle \frac{c - c}{a} = 0, and \overline{BC} lies on the x-axis.

Figure 14Rows 1 to 9 of the table. BC = 2a uses the placement and nothing else.

The flowchart of rows 1 to 9 shows what the table hides: BC = 2a has an arrow only from the placement and never uses a midpoint. Rows 10 and 11 hang from rows 3 and 4 as a separate branch (Exercise 20), so the two halves of the Prove are two proofs sharing a set-up. A synthetic proof needs similar triangles or a parallelogram built on \overline{DE}, both later on the map (the midsegment theorem); the coordinate proof of the midsegment theorem treats this route in full.

Example 10

A placement that proves too little

One student proves the midsegment theorem with B(0, 0), C(2a, 0) and A(0, 2c). Another proves Example 6 with A(0, 0), B(a, 0), C(a, a) and D(0, a). Every computation is right. What has each proved?

Solution

The first placement puts A straight above B, so \angle ABC is right. The work gives D(0, c), E(a, c) and DE = a = \tfrac{1}{2}BC, but it is Example 9 with b = 0: a proof for right triangles only. The second placement makes AB = AD = a, so it proves the theorem for squares only.

The test is to count free letters: a triangle with one side on the axis needs three, a rectangle two. Fewer letters, or one letter in two places the Given does not tie together, proves a special case; numbers in place of letters check a single figure.

Boundaries

A change of form cannot mend a wrong step. A reason that does not fit is wrong in a table, a sentence and a box alike. A flowchart shows which steps depend on which, not that the steps are right, and an arrow from the wrong box hides a gap.

The arrows of a flowchart may not loop. If box P leads to box Q and Q leads back to P, each has been used to prove the other. That is circular reasoning, and no order of writing puts each after the boxes it needs.

A coordinate proof is only as general as its placement (Example 10), and its tools must come before its theorem. The distance formula rests on the Pythagorean theorem, so it cannot be used to prove that theorem (Exercise 33).

Distance, midpoint and slope reach lengths, not angle measures. An angle of 40^\circ has no simple form in coordinates, so theorems about angle measures are usually proved synthetically.

Some arguments fit a flowchart badly. A long proof with many joins makes arrows cross, and the chart then serves better as a plan than as the finished proof. A proof by cases runs one argument several times, and an indirect proof makes an assumption it later throws away; a chart has no natural box for either, so those are usually written as paragraphs or in two columns.

Where this goes next

Choosing a general placement is a skill of its own, in placing a figure on the coordinate plane. Flowcharts are at their best where separate facts join, as they do in many angle proofs such as proving vertical angles congruent and proofs with parallel lines, and in the congruence proofs of planning a congruence proof, where three branches, one for each pair of parts, meet at one congruence. The synthetic routes of Examples 8 and 9 open up with congruent triangles and the midsegment theorem, and the converse of Example 6, that a parallelogram with congruent diagonals is a rectangle, is proving a parallelogram is a rectangle. Boxes joined by one-way arrows make a directed graph, and the orders a chart allows are the orders in which any set of tasks can be done when some must come before others. Beyond school geometry nearly every proof is a paragraph, but the habit the two-column form trains, a reason for every step, is what deductive reasoning asks of all of them.

Exercises

Reading a flowchart

How many orders could a two-column proof of the vertical angles theorem take, going by the flowchart drawn under the Definition?
In the flowchart of Example 3, which boxes could trade places with m\angle 1 + m\angle 2 + m\angle 3 = 180^\circ in a two-column proof? Why?
In the flowchart of Example 4, name the two boxes that have two arrows leaving them.
How many orders could a two-column proof of Example 2 take? List them.

Numbers through a flowchart

In Example 3, m\angle 1 = 52^\circ and m\angle 2 = 71^\circ. Find m\angle 3 and m\angle 4 from the boxes that give them, and check the Prove.
In Example 1, m\angle ABD = (3x + 6)^\circ and m\angle DBC = (5x - 4)^\circ. Use the box m\angle ABD + m\angle DBC = 90^\circ to find x and both angles.
In Example 4, m\angle 1 = (4x + 10)^\circ and m\angle 2 = (6x - 30)^\circ. Use the box m\angle 1 = m\angle 2 to find x, and check the result against the Prove.

From two columns to a flowchart

Exercises 8 to 10 use this proof. Given: \angle 1 and \angle 3 are supplementary; \angle 2 and \angle 3 are supplementary. Prove: \angle 1 \cong \angle 2. Its reasons have been left without the line numbers they use; supplying them is Exercise 8.

Statements	Reasons
1. \angle 1 and \angle 3 are supplementary	given
2. m\angle 1 + m\angle 3 = 180^\circ	definition of supplementary angles
3. \angle 2 and \angle 3 are supplementary	given
4. m\angle 2 + m\angle 3 = 180^\circ	definition of supplementary angles
5. 180^\circ = m\angle 2 + m\angle 3	symmetric property of equality
6. m\angle 1 + m\angle 3 = m\angle 2 + m\angle 3	substitution
7. m\angle 1 = m\angle 2	subtraction property of equality
8. \angle 1 \cong \angle 2	definition of congruent angles

For each of statements 2 to 8, list the earlier statements its reason uses.
Sketch the flowchart. Which boxes sit side by side, and where do the branches join?
Could statement 3 be written first? Could statement 4 be written before statement 2? Could statement 6 be written before statement 5?

From a flowchart to two columns

Write the flowchart of Example 1 as a two-column proof, with each reason naming the lines it uses. How many orders of the lines are possible?

Building a flowchart

Draw a flowchart proof. The answers list each box with its reason and the boxes pointing into it.

Given: \overrightarrow{BD} bisects \angle ABC. Prove: m\angle ABD = \tfrac{1}{2}m\angle ABC.
Given: l \parallel m, angles numbered as in Example 2. Prove: \angle 4 \cong \angle 5.
Given: l \parallel m, angles numbered as in Example 2. Prove: \angle 3 and \angle 5 are supplementary.
Given: \angle 1 and \angle 2 are vertical angles; \angle 2 and \angle 3 form a linear pair. Prove: \angle 1 and \angle 3 are supplementary.

Translating between forms

Exercises 16 to 18 use this paragraph proof. Given. B is the midpoint of \overline{AC}; M is the midpoint of \overline{AB}. Prove. AM = \tfrac{1}{4}AC. “Because M is the midpoint of \overline{AB}, AM = MB, and since AM + MB = AB, AB = 2AM. In the same way AC = 2AB. Substituting, AC = 4AM, so AM = \tfrac{1}{4}AC.”

Write the proof in two columns, with a reason on every row. Which steps did “in the same way” leave out?
For each row of your answer to Exercise 16, list the rows its reason uses. Where does the flowchart branch, and where do the branches join?
In the same figure, AC = 36. Find AB, AM and MC.
Write rows 1 to 9 of Example 9 as a paragraph proof.
Add boxes for rows 10 and 11 of Example 9 to its flowchart. Which boxes send arrows into them?

Computing with coordinates

A rectangle has vertices (1, 2), (9, 2), (9, 8) and (1, 8). Find the length of each diagonal.
A triangle has vertices B(0, 0), C(10, 0) and A(4, 6). Find the midpoints D of \overline{AB} and E of \overline{AC}, and check both parts of the midsegment theorem.
A triangle has vertices A(1, 7), B(-3, 1) and C(5, -1). With D and E the midpoints of \overline{AB} and \overline{AC}, find DE and BC, and compare the slopes of \overline{DE} and \overline{BC}.
Using the placement of Example 6, find the midpoints of \overline{AC} and \overline{BD}. What theorem have you proved?

Placements

For “the diagonals of a rectangle are congruent”, say whether each placement is general: (a) A(0, 0), B(a, 0), C(a, b), D(0, b); (b) A(0, 0), B(2a, 0), C(2a, 2a), D(0, 2a); (c) A(-a, -b), B(a, -b), C(a, b), D(-a, b); (d) A(0, 0), B(8, 0), C(8, 6), D(0, 6).
Redo rows 4 to 6 of Example 9 with C(a, 0) and A(b, c) instead. What changes, and what does not?
A student proves the midsegment theorem with B(0, 0), C(2a, 0) and A(2a, 2c). Which triangles does the proof cover?

Choosing a form

Say which form you would choose for each, and why: (a) the diagonals of a square are perpendicular; (b) if \overrightarrow{BD} bisects a right angle \angle ABC, then m\angle ABD = 45^\circ; (c) a triangle cannot have two right angles.

Reasoning

A chart for Example 2 has an extra box, \angle 1 \cong \angle 5 by the Corresponding Angles Theorem, with an arrow into it from l \parallel m and no arrow leaving it. Is the proof wrong? What should happen to the box?
Explain why two boxes with no path of arrows between them can be written in either order in a two-column proof, while a box must come after every box that has a path of arrows into it.
A flowchart has an arrow from box P to box Q and another from Q back to P. Explain what is wrong with the argument.
Explain why a two-column proof and a flowchart built from the same statements and reasons are either both correct or both wrong.
A student proves the Pythagorean theorem by placing a right triangle at (0, 0), (a, 0) and (0, b) and finding the hypotenuse with the distance formula. What is wrong?

Answers

3. The Given comes first, since both boxes under it need it. Then come the two linear pair equations and the symmetric box, which must follow the equation it turns around, so the first equation goes before, between or after the other two. Then the last three boxes, in their one possible order.
m\angle 3 + m\angle 4 = 180^\circ and 180^\circ = m\angle 3 + m\angle 4. The triangle sum box needs only the Given, and there is no path of arrows between it and either of those two, so it can go before, between or after them. Those two cannot trade with each other, since the symmetric box uses the linear pair box above it.
m\angle 2 = m\angle 1, used for m\angle 1 + m\angle 1 = 180^\circ and again for m\angle 2 = 90^\circ; and m\angle 1 = 90^\circ, used for m\angle 2 = 90^\circ and for the last box.
3. With G for l \parallel m, A for \angle 2 \cong \angle 6, B for \angle 3 \cong \angle 2 and C for the Prove: G A B C, G B A C and B G A C. C comes last and A after G, but B needs nothing, so it can take any of the first three places.
m\angle 3 = 57^\circ and m\angle 4 = 123^\circ. The triangle angle sum box gives m\angle 3 = 180^\circ - 52^\circ - 71^\circ = 57^\circ, the linear pair box gives m\angle 4 = 180^\circ - 57^\circ = 123^\circ, and 52^\circ + 71^\circ = 123^\circ, as the Prove says.
x = 11, m\angle ABD = 39^\circ and m\angle DBC = 51^\circ. The box gives (3x + 6) + (5x - 4) = 90, so 8x + 2 = 90 and 8x = 88. Check: 39 + 51 = 90.
x = 20, and both angles are 90^\circ. The box gives 4x + 10 = 6x - 30, so 2x = 40. Then 4(20) + 10 = 90 and 6(20) - 30 = 90: right angles, as the Prove says.
2 uses 1; 4 uses 3; 5 uses 4; 6 uses 2 and 5; 7 uses 6; 8 uses 7.
The givens 1 and 3 sit side by side at the top, each with an arrow down to its own equation, 2 and 4, which also sit side by side; the right branch runs on from 4 to 5. The branches join at 6, which has two arrows in; then a single line runs through 7 to 8.
Yes: 3 is a given and needs nothing. Yes: 4 needs only 3, not 2. No: 6 uses 5, so 5 must be above it.
One order: (1) \angle ABC is a right angle (given). (2) D is inside \angle ABC (given). (3) m\angle ABC = 90^\circ (definition of right angle: line 1). (4) m\angle ABD + m\angle DBC = m\angle ABC (angle addition postulate: D is inside \angle ABC, line 2). (5) m\angle ABD + m\angle DBC = 90^\circ (substitution: line 3 into line 4). (6) \angle ABD and \angle DBC are complementary (definition of complementary angles: line 5). The last two are fixed; the first four may go in any order that keeps each given above its own equation, which is \binom{4}{2} = 6 orders, counted as in Example 5.
\overrightarrow{BD} bisects \angle ABC, given. m\angle ABD = m\angle DBC, definition of angle bisector, from the given. m\angle ABD + m\angle DBC = m\angle ABC, angle addition postulate, from the given, since a bisector lies inside its angle. m\angle DBC = m\angle ABD, symmetric property of equality, from the bisector equation. m\angle ABD + m\angle ABD = m\angle ABC, substitution, from the symmetric box and the angle addition box. 2\,m\angle ABD = m\angle ABC, simplify. m\angle ABD = \tfrac{1}{2}m\angle ABC, division property of equality. One given, two branches (one of them two boxes long), a join, then a single line.
l \parallel m, given. \angle 4 \cong \angle 8, Corresponding Angles Theorem, from l \parallel m. \angle 8 \cong \angle 5, Vertical Angles Theorem, no arrow in. \angle 4 \cong \angle 5, transitive property of congruence, from both. The same shape as Example 2.
l \parallel m, given. \angle 1 \cong \angle 5, Corresponding Angles Theorem, from l \parallel m. m\angle 1 = m\angle 5, definition of congruent angles, from that. m\angle 1 + m\angle 3 = 180^\circ, Linear Pair Theorem, no arrow in: \angle 1 and \angle 3 share the ray along l to the left, and their other sides run opposite ways along t. m\angle 5 + m\angle 3 = 180^\circ, substitution, from the two equations. \angle 3 and \angle 5 are supplementary, definition of supplementary angles.
Two givens. \angle 2 \cong \angle 1, Vertical Angles Theorem, from the first given. m\angle 2 = m\angle 1, definition of congruent angles, from that. m\angle 2 + m\angle 3 = 180^\circ, Linear Pair Theorem, from the second given. m\angle 1 + m\angle 3 = 180^\circ, substitution, from the two equations: m\angle 1 in place of m\angle 2. \angle 1 and \angle 3 are supplementary, definition of supplementary angles. Two branches, one of them two boxes long, join at the substitution.
(1) M is the midpoint of \overline{AB}, given. (2) AM = MB, definition of midpoint: line 1. (3) MB = AM, symmetric property of equality: line 2. (4) AM + MB = AB, segment addition postulate: M is between A and B, line 1. (5) AM + AM = AB, substitution: line 3 into line 4. (6) 2AM = AB, simplify: line 5. (7) AB = 2AM, symmetric property of equality: line 6. (8) B is the midpoint of \overline{AC}, given. (9) AB = BC, definition of midpoint: line 8. (10) BC = AB, symmetric property of equality: line 9. (11) AB + BC = AC, segment addition postulate: B is between A and C, line 8. (12) AB + AB = AC, substitution: line 10 into line 11. (13) 2AB = AC, simplify: line 12. (14) 2(2AM) = AC, substitution: line 7 into line 13. (15) 4AM = AC, simplify: line 14. (16) AM = \tfrac{1}{4}AC, division property of equality: line 15. “In the same way” stood for rows 8 to 12, and never said that B is a given.
2 and 4 use 1; 3 uses 2; 5 uses 3 and 4; 6 uses 5; 7 uses 6; 9 and 11 use 8; 10 uses 9; 12 uses 10 and 11; 13 uses 12; 14 uses 7 and 13; 15 uses 14; 16 uses 15. Each given starts a branch that splits in two and joins again (at 5 and at 12), and the two branches join at 14.
AB = 18, AM = 9 and MC = 27. B halves \overline{AC} and M halves \overline{AB}, so AM = \tfrac{1}{4}(36) = 9, and MC = 36 - 9 = 27.
Place B at the origin, C at (2a, 0) and A at (2b, 2c). By the midpoint formula D = (b, c) and E = (a + b, c), so by the distance formula DE = a. Also by the distance formula BC = 2a. Dividing by 2 gives \tfrac{1}{2}BC = a, that is a = \tfrac{1}{2}BC, and substituting \tfrac{1}{2}BC for a in DE = a gives DE = \tfrac{1}{2}BC.
A box “slopes of \overline{DE} and \overline{BC} are 0”, reason slope formula, with arrows from D(b, c), E(a + b, c) and from the placement; under it a second double box, \overline{DE} \parallel \overline{BC}, reason equal slopes, with arrows from the slopes box and from the placement, which supplies c > 0. No arrow joins this branch to the boxes DE = a onwards.
Both are 10: \displaystyle \sqrt{(9 - 1)^2 + (8 - 2)^2} = \sqrt{100} from (1, 2) to (9, 8), and \displaystyle \sqrt{(1 - 9)^2 + (8 - 2)^2} = \sqrt{100} from (9, 2) to (1, 8).
D(2, 3) and E(7, 3). DE = 7 - 2 = 5 and BC = 10, so DE = \tfrac{1}{2}BC. Both slopes are 0, and \overline{DE} sits at height 3, off the x-axis, so \overline{DE} \parallel \overline{BC}.
D(-1, 4) and E(3, 3). \displaystyle DE = \sqrt{4^2 + (-1)^2} = \sqrt{17} and \displaystyle BC = \sqrt{8^2 + (-2)^2} = \sqrt{68} = 2\sqrt{17}, so DE = \tfrac{1}{2}BC. Both slopes are -\tfrac{1}{4}: \displaystyle \frac{3 - 4}{3 - (-1)} and \displaystyle \frac{-1 - 1}{5 - (-3)}. D is not on line BC, so the segments are parallel.
Both are \left(\tfrac{a}{2}, \tfrac{b}{2}\right): \displaystyle \left(\frac{0 + a}{2}, \frac{0 + b}{2}\right) and \displaystyle \left(\frac{a + 0}{2}, \frac{0 + b}{2}\right). The diagonals of a rectangle cross at the midpoint of each, so they bisect each other.
(a) General: a and b are any positive lengths. (b) Squares only: both sides are 2a. (c) General: sides 2a and 2b, any positive values, centered on the origin, where any rectangle can be slid. (d) One 8 by 6 rectangle: a check, not a proof.
D = \left(\tfrac{b}{2}, \tfrac{c}{2}\right), E = \left(\tfrac{a + b}{2}, \tfrac{c}{2}\right), DE = \tfrac{a + b}{2} - \tfrac{b}{2} = \tfrac{a}{2} and BC = a. The result is the same, DE = \tfrac{1}{2}BC, and the placement is just as general; only the fractions are new.
Only triangles with a right angle at C, since A sits straight above C. It is Example 9 with b = a, a special case.
(a) Coordinates: with A(0, 0), B(a, 0), C(a, a), D(0, a) the diagonals have slopes 1 and -1, whose product is -1 (perpendicular slopes); a synthetic proof would need congruent triangles. (b) Synthetic, in two columns or a sentence: a right angle is 90^\circ and the bisector halves it. Coordinates reach angle measures badly. (c) An indirect proof in a paragraph: two right angles would leave 0^\circ for the third angle.
Not wrong: the box is true and its reason is right. But no path of arrows leads from it to the Prove, so nothing uses it. It belongs in the working, and the finished proof leaves it out.
A statement can be written once everything its reason uses is above it. With no path between two boxes, neither is used by the other, even through other boxes, so either can go first. A path from P to Q means Q uses something that, step by step, uses P, so P must be written first.
P is used to prove Q and Q to prove P, so each rests on the other and neither rests on the Given. That is circular reasoning. No two-column order works either, since each of the two would have to come before the other.
A proof is correct when every reason fits its statement and uses only statements already established. The same reasons fit or fail the same statements in both forms. And “above it” in the table and “a path of arrows into it” in the chart agree for every order the chart allows, while a chart with a loop has no such order at all.
The distance formula is the Pythagorean theorem applied between two points, so the proof uses the theorem to prove itself. That is circular.