Converses and biconditional definitions
A theorem says that if one thing is true of a figure, then another is. Once it is proved, what do you know about the statement that runs the other way, and why may a definition always be run backwards when a theorem may not?
Where you are 18 in Geometric proof
Why this exists
Every line of a proof needs a reason, and the reasons come in kinds that behave differently when you turn them around.
The most common reason in a geometry proof is a definition. Told that M is the midpoint of \overline{AB}, you write AM = MB, reason “definition of midpoint”. Shown that a point M on \overline{AB} has AM = MB, you write “M is the midpoint of \overline{AB}”, and the reason is again “definition of midpoint”. The same reason runs in both directions.
Theorems do not behave like that. Two theorems from the same chapter of geometry look alike:
- If two angles are vertical angles, then they are congruent.
- If two lines cut by a transversal are parallel, then their corresponding angles are congruent.
Turn each one around. “If two angles are congruent, then they are vertical angles” is false: two 30^\circ angles drawn in different corners of a page are congruent and nowhere near each other. “If corresponding angles are congruent, then the lines are parallel” is true, and it is the standard way to show that two lines are parallel.
Same shape of sentence, opposite fates. So a theorem and its converse are two claims, each needing its own proof, and the converses that come back true become geometry’s tests. A definition runs both ways for a different reason: it is not a fact someone found out but an agreement about what a word will mean, and the agreement covers both directions at once.
Definition
Definitions, postulates and theorems
Apart from what is given and the properties of equality (the rules of algebra), a proof’s reasons come in three kinds.
- A definition says what a word means, by naming the exact condition the word stands for, like the definition of midpoint above. It makes no claim, so it is never proved.
- A postulate, also called an axiom, is a statement accepted without proof, like the segment addition postulate: if B is between A and C, then AB + BC = AC. Postulates are where the chain of reasons starts; see axioms, definitions and theorems.
- A theorem is a statement proved from definitions, postulates and earlier theorems, like the Vertical Angles Theorem.
Underneath the definitions sit the undefined terms, point, line and plane, which every other word is defined from.
One more source is the figure itself. Given, prove and the diagram sets out what a diagram may be trusted for: which points lie on which lines and in what order, but not lengths, angle sizes, right angles or parallels. Those must be given, marked or proved.
The three kinds differ when a reason is run backwards. A definition runs backwards for free. A postulate and a theorem each claim exactly what they say, so the statement turned around is a new claim that needs its own support. The rest of this lesson is about those two cases.
A theorem and its converse
Most theorems in geometry are conditionals about a kind of figure. The setting names the figure (a triangle ABC; two lines cut by a transversal), the hypothesis is what is assumed about it, and the conclusion is what is proved. The converse keeps the setting and swaps the other two.
| Setting | two lines cut by a transversal |
|---|---|
| theorem | if the lines are parallel, then corresponding angles are congruent |
| converse | if corresponding angles are congruent, then the lines are parallel |
The setting belongs to neither claim; it is what both claims are about.
The logic of turning a conditional around is worked out in the contrapositive. This lesson needs one fact from it: a conditional and its converse are not logically equivalent. Either can be true while the other is false.
When a theorem and its converse are both true, they combine into one biconditional, “hypothesis if and only if conclusion”, and each condition is necessary and sufficient for the other.
Why definitions work both ways
A definition gives a word its meaning by naming a condition, and it says the word applies to exactly the things that meet the condition: all of them, and nothing else. So every definition is a biconditional. For a word W defined by a condition C, and any object x,
Take the definition of perpendicular lines: two lines are perpendicular if they meet to form a right angle. In full it is two conditionals.
- Forward: if two lines are perpendicular, then they meet to form a right angle.
- Backward: if two lines meet to form a right angle, then they are perpendicular.
The forward direction unpacks the word: it turns a name you were handed into a fact you can use. The backward direction earns the word: once the condition is checked, the name is yours.
Definitions are usually written with a plain “if”, as above, and read as “if and only if”. The shorter wording is a habit of style, not a weaker claim, and it is kept only for definitions. A theorem claims exactly what it says.
Intuition
Two groups of figures
Picture every figure in the setting. Some satisfy the hypothesis and some satisfy the conclusion. A true theorem says the first group sits inside the second. The converse says the second sits inside the first. Both at once means the two groups are the same group.
Vertical pairs are a small corner of all congruent pairs, so that converse fails. For parallel lines the two groups are identical. A proof of the theorem looks only at the first group, so it cannot tell you which case you are in.
A definition as a test at the door
A definition is the case where the two groups are the same by decree. Think of it as a test at a door. Everything the word is meant to name must pass, and everything that passes must be something the word is meant to name. Each direction of the biconditional guards one half of that, so a proposed test can fail in two ways.
It is too loose if it lets in something it should not: the backward direction is false, and the counterexample meets the condition but is not the thing. It is too tight if it turns away something it should let in: the forward direction is false, and the counterexample is a genuine example that fails the condition. One counterexample on either side sinks it.
Results
Theorems and their converses
A theorem does not prove its converse. Each direction is a separate statement, and needs its own proof or its own counterexample.
One counterexample settles a false converse. It is a figure in the setting where the conclusion holds and the hypothesis fails.
Both directions proved give an if and only if. Some theorems of geometry, each proved in its own lesson, and their converses:
| Theorem | Converse |
|---|---|
| vertical angles are congruent | false |
| a linear pair is supplementary | false |
| if l \parallel m, corresponding angles are congruent | true |
| if l \parallel m, alternate interior angles are congruent | true |
| if l \parallel m, same side interior angles are supplementary | true |
| if m\angle C = 90^\circ in \triangle ABC, then a^2 + b^2 = c^2 | true |
| if AB = AC in \triangle ABC, then \angle B \cong \angle C | true |
| a rectangle’s diagonals are congruent | false |
Each “true” row is a biconditional. For two lines cut by a transversal, with corresponding angles \angle 2 and \angle 6 as in Figure 1 below,
and in a triangle whose sides a, b, c lie opposite \angle A, \angle B, \angle C,
Definitions
A definition works in both directions. If W is defined by condition C, then
so “definition of W” is a valid reason for a step in either direction.
Testing a proposed definition. A condition C captures an idea exactly when every example of the idea meets C (not too tight) and everything that meets C is an example (not too loose).
A theorem with a false converse cannot be a definition. If every W has property P but something that is not a W has it too, then P is a true fact about W and does not define it.
Equivalent definitions. If conditions C_1 and C_2 hold for exactly the same objects, either one can serve as the definition, and the other then becomes a theorem.
Why it is true
Why a definition runs both ways
Before a word is defined it has no meaning in the system, so there is nothing to discover about it. The definition declares that W names the objects meeting C and no others. “Names the objects meeting C” is the backward direction. “And no others” is the forward direction: nothing that fails C is a W. Neither direction is proved, because both were agreed at the start, and neither can be disproved, because there is no other source for what W means.
A theorem joins two conditions that already have meanings. Vertical angles are defined by position and then proved equal in measure. Equal measure is a separate idea, so it can turn up without that position, as Example 8 shows.
The test of a proposed definition reads W \iff C one direction at a time, so it fails when either direction fails. For equivalent definitions, W \iff C_1 and C_1 \iff C_2 give W \iff C_2.
How true converses are proved
That a theorem does not prove its converse, and that one counterexample settles a false one, is logic (A conditional and its contrapositive). The interest is in how the true converses are proved: in each, the two directions need different arguments. Euclid’s Elements proves them as separate propositions of Book I: 5 and 6 for the isosceles triangle, 29 against 27 and 28 for parallels, 47 and 48 for Pythagoras. Facts borrowed from later lessons are linked where they are used.
Parallel lines, the converse. In Figure 1, lines l and m are cut by t at P and Q. Suppose \angle 2 \cong \angle 6. If l and m met at some point X to the right of t, then P, Q and X would form a triangle. Its angle at Q is \angle 6. Its angle at P is \angle 4, which forms a linear pair with \angle 2. So
But any two angles of a triangle add to less than 180^\circ. That is Euclid’s Proposition 17, which comes from Proposition 16 (an exterior angle is larger than either far interior angle) and uses no parallel postulate, as The parallel postulate explains. Quoting the Triangle angle sum here instead would go round in a circle, since that theorem is itself proved from parallel lines. If the lines met to the left of t, the same argument runs with \angle 3 and \angle 5. So they never meet, and l \parallel m.
Parallel lines, the theorem. Now suppose l \parallel m. Through P draw the line l' whose angle with t, in the position of \angle 2, equals \angle 6. By the converse just proved, l' \parallel m. But through P there is only one line parallel to m; that is the parallel postulate. So l' is l, and \angle 2 \cong \angle 6.
The two directions stand on different ground: the converse needs only Proposition 17, the theorem needs the postulate. This is Euclid’s route, also told in The parallel postulate. The map’s Corresponding angles lesson proves both directions with a slide of the plane instead, and Alternate and same side angles follows from it; there too it is the theorem direction that calls on the postulate. The full treatment is Proving lines parallel.
Pythagoras, the converse. Take \triangle ABC with a^2 + b^2 = c^2 (Figure 2, left). Build \triangle DEF with a right angle at F and legs EF = a, FD = b. By the Pythagorean theorem, DE^2 = a^2 + b^2 = c^2, so DE = c. The triangles match side for side. SSS, one of the congruence criteria set out in a later lesson, will make them congruent, and then \angle C \cong \angle F, a right angle. So the plan proves the converse by using the theorem, on a different triangle; the proof itself waits for SSS, in Converse of the Pythagorean theorem.
Worked examples
Both directions of one definition
\overrightarrow{BD} bisects \angle ABC when D is in the interior of \angle ABC and \angle ABD \cong \angle DBC (the angle bisector). Write this as two conditionals, then say which one each part uses.
(a) \overrightarrow{BD} bisects \angle ABC and m\angle ABD = 34^\circ. Find m\angle DBC and m\angle ABC.
(b) D is in the interior of \angle ABC, m\angle ABD = (3x + 4)^\circ, m\angle DBC = (5x - 12)^\circ and m\angle ABC = 56^\circ. Does \overrightarrow{BD} bisect \angle ABC?
SolutionForward: if \overrightarrow{BD} bisects \angle ABC, then D is in its interior and \angle ABD \cong \angle DBC. Backward: if D is in the interior and \angle ABD \cong \angle DBC, then \overrightarrow{BD} bisects \angle ABC.
(a) Here the name is given, so the forward direction unpacks it: \angle DBC \cong \angle ABD, so m\angle DBC = 34^\circ. It also puts D in the interior, so the angle addition postulate applies: m\angle ABC = 34^\circ + 34^\circ = 68^\circ.
(b) Here the name is what we want, so we must check the condition. D is in the interior, so
Then m\angle ABD = 3(8) + 4 = 28^\circ and m\angle DBC = 5(8) - 12 = 28^\circ. Equal measures make the angles congruent (the definition of congruent angles, backward), and with D inside, the backward direction of the bisector definition says yes, it bisects.
Both parts cite “definition of angle bisector”. Where you start, from the name or from the condition, sets which direction it runs.
Sorting the reasons in a proof
(a) M is the midpoint of \overline{AB} and AB = 18. Find AM, giving a reason for each step and saying what kind of reason it is.
(b) Lines \ell and n cross at O, \angle 1 and \angle 3 are vertical angles there, and \angle 1 is a right angle. Show that \angle 3 is a right angle and that \ell \perp n, again naming the kind of each reason.
Solution(a)
| Statement | Reason | Kind |
|---|---|---|
| M is the midpoint of \overline{AB} | given | given |
| AB = 18 | given | given |
| M lies on \overline{AB} | definition of midpoint | definition, forward |
| AM = MB | definition of midpoint | definition, forward |
| AM + MB = AB | segment addition postulate | postulate |
| AM + AM = 18 | substitution | property of equality |
| AM = 9 | division by 2 | property of equality |
The postulate needs M between A and B. The definition supplies it: M is on the segment, and it is neither end, since M = A would make MB = AM = 0 and A and B the same point.
(b)
| Statement | Reason | Kind |
|---|---|---|
| \angle 1 and \angle 3 are vertical | given | given |
| \angle 1 is a right angle | given | given |
| m\angle 1 = 90^\circ | definition of right angle | definition, forward |
| \angle 1 \cong \angle 3 | Vertical Angles Theorem | theorem |
| m\angle 3 = m\angle 1 | definition of congruent angles | definition, forward |
| m\angle 3 = 90^\circ | substitution | property of equality |
| \angle 3 is a right angle | definition of right angle | definition, backward |
| \ell \perp n | definition of perpendicular lines | definition, backward |
The definitions ran in both directions, sometimes the same one twice. The theorem ran only in the direction it was proved: from vertical to congruent, never back.
A converse that fails inside one figure
Write the converse of the Vertical Angles Theorem and decide whether it is true.
The theorem is “if two angles are vertical angles, then they are congruent.” Swapping hypothesis and conclusion gives the converse: “if two angles are congruent, then they are vertical angles.”
To refute it, find two angles that are congruent (the new hypothesis) and not vertical (the new conclusion fails). In Figure 3 the lines are perpendicular, so m\angle 1 = m\angle 2 = 90^\circ and \angle 1 \cong \angle 2. But the two share a side: they are a linear pair, not a vertical pair. The converse is false.
Which direction does the problem need?
Lines l and m are cut by a transversal t. What can you conclude, and from which statement?
(a) A pair of corresponding angles measure 72^\circ and 72^\circ.
(b) l \parallel m, and one angle at the upper crossing measures 72^\circ.
(c) A pair of alternate interior angles measure 72^\circ and 70^\circ.
Solution(a) The facts given are about angles and the conclusion wanted is about lines, which is the direction of the converse. Corresponding angles are congruent, so l \parallel m.
(b) The lines are given parallel, so the theorem applies: the corresponding angle at the lower crossing measures 72^\circ as well, and its neighbors 180^\circ - 72^\circ = 108^\circ.
(c) The Alternate Interior Angles Theorem says: if l \parallel m, these angles are congruent. They are not, so the conclusion fails, and then the hypothesis must fail too: l and m are not parallel. That step is the theorem’s contrapositive, not its converse.
The value that makes two lines parallel
In Figure 4, find the value of x for which l \parallel m.
\angle 3 and \angle 6 are alternate interior angles. By the converse, the lines are parallel when the two are congruent:
Then 4(21) - 7 = 77 and 2(21) + 35 = 77, so each angle is 77^\circ.
For any other x the angles differ, and by the theorem’s contrapositive the lines are not parallel. So l \parallel m if and only if x = 21. The converse alone shows that 21 works; the theorem is what shows that nothing else does.
Testing triangles for a right angle
Is either triangle a right triangle? (a) sides 20, 21, 29; (b) sides 6, 9, 11.
SolutionIn a right triangle c^2 = a^2 + b^2, so the hypotenuse is the longest side. Any right angle must sit opposite the longest side, so test that one.
(a) 20^2 + 21^2 = 400 + 441 = 841 = 29^2. By the converse of the Pythagorean theorem, the angle opposite the side of 29 is a right angle.
(b) 6^2 + 9^2 = 36 + 81 = 117, while 11^2 = 121. If the angle opposite the side of 11 were right, the theorem would make these equal; they are not. No other angle can be right, so the triangle is not right.
Part (a) needed the converse and part (b) needed the theorem.
A definition that lets too much in
Someone proposes: “Two lines are parallel if they do not intersect.” Test both directions.
SolutionForward: parallel lines are, by their true definition, lines that lie in one plane and do not meet. So they do not intersect, and this direction holds.
Backward: must two lines that do not intersect be parallel? Draw a straight line across a table top, and hold a pencil level a hand’s width above the table, pointing across that line at an angle. Extend the pencil’s line both ways. It stays a hand’s width up, so it never meets the line on the table, yet the two run in different directions. Two lines in one plane that never meet would run the same way, so no flat sheet holds both. They are skew, not parallel, and the proposal is too loose.
Repair it by adding what the skew pair lacks: “Two lines are parallel if they lie in one plane and do not intersect.” The forward direction still holds, and the pencil and the drawn line now fail the test. The clause “in one plane” is there only because of the backward direction.
A true theorem that cannot be a definition
Someone proposes: “Two angles are vertical angles if they share a vertex and are congruent.” Test both directions, using Figure 5.
Forward: vertical angles share their vertex by definition, and they are congruent by the Vertical Angles Theorem. This direction holds.
Backward: in Figure 5, \overrightarrow{BD} splits a 100^\circ angle into two 50^\circ angles. \angle ABD and \angle DBC share the vertex B and are congruent, so they meet the condition. But they share a whole side, \overrightarrow{BD}, and sit side by side; they are not opposite each other at a crossing. The proposal is too loose.
The forward direction was a true theorem, and that is the trap: a true fact about vertical angles need not pick out vertical angles alone. They are defined by position (their sides form two pairs of opposite rays), and congruence is proved from that. This is Example 3’s false converse seen from the other side.
A definition that fails both ways
Someone proposes: “A line bisects \overline{AB} if it is perpendicular to \overline{AB}.” Test both directions, using Figure 6, where M is the midpoint of \overline{AB}, line \ell passes through M at 60^\circ to \overline{AB}, and line n is perpendicular to \overline{AB} at a point P other than M.
Forward: must every bisector be perpendicular? Line \ell passes through the midpoint, so it is a segment bisector of \overline{AB}, but it meets it at 60^\circ, not 90^\circ. It is a real bisector that fails the condition, so the proposal is too tight.
Backward: must every perpendicular bisect? Line n meets \overline{AB} at a right angle, but at P, and a segment has only one midpoint, M. So n meets the condition and is not a bisector, and the proposal is too loose as well.
Perpendicular and through the midpoint together describe one particular bisector, the perpendicular bisector, not all of them.
Planning a converse: equal base angles
The Base Angles Theorem says: in \triangle ABC, if AB = AC, then \angle B \cong \angle C. Plan a proof of the converse, and state the two together.
The converse is: in \triangle ABC, if \angle B \cong \angle C, then AB = AC. The theorem cannot be quoted: it assumes the very sides we want to prove equal.
Every angle has an angle bisector, and it can be drawn with compass and straightedge (Bisecting an angle). The bisector of \angle A runs into the inside of the angle, so it crosses the opposite side; call the crossing point D, as in Figure 7. Compare \triangle ABD and \triangle ACD:
- \angle B \cong \angle C, given;
- \angle BAD \cong \angle CAD, because \overline{AD} bisects \angle A (definition of angle bisector, forward);
- AD = AD, a shared side, opposite \angle B in one triangle and \angle C in the other.
Two angles and a side not between them match. That is the if-part of AAS, one of the congruence criteria proved in a later lesson, which will make the triangles congruent and so their corresponding sides equal: AB = AC. Until AAS is proved this is a plan, not a proof; the theorem has its own page, Converse of the base angles theorem, and this bisector proof is written out in full as Exercise 17 of proving the Base Angles Theorem.
Together, once both are proved: in \triangle ABC, AB = AC if and only if \angle B \cong \angle C. The plan shows the shape of the lesson: the converse needs a different argument from the theorem, and here a tool the theorem did not.
Repairing a false converse
A square’s diagonals are congruent and perpendicular. Is the converse true? If not, strengthen its hypothesis until it is.
The converse is: if a quadrilateral’s diagonals are congruent and perpendicular, then it is a square. Try the kite with A(-2, 0), B(0, 1), C(2, 0), D(0, -3) in Figure 8.
Diagonal \overline{AC} lies on the x-axis with length 4, and \overline{BD} on the y-axis with length 1 + 3 = 4: congruent, and perpendicular because the axes are. But the sides are
so it is not even a rhombus, let alone a square. The converse is false.
What the kite lacks: O cuts \overline{AC} in half but not \overline{BD}. Add that. Diagonals that bisect each other make a parallelogram (Proving a quadrilateral is a parallelogram); congruent diagonals then make it a rectangle, and perpendicular ones a rhombus, so it is a square. A square’s diagonals do bisect each other, so: a quadrilateral is a square if and only if its diagonals are congruent, perpendicular, and bisect each other.
Two definitions that say the same thing
For a ray \overrightarrow{BD} and an angle \angle ABC, call this the half condition: D is in the interior of \angle ABC and m\angle ABD = \tfrac{1}{2}\, m\angle ABC. Show that the half condition holds exactly when \overrightarrow{BD} bisects \angle ABC, so either could serve as the definition. Then use Figure 9 to show the interior clause cannot be dropped.
Both conditions put D in the interior, so the angle addition postulate gives m\angle ABD + m\angle DBC = m\angle ABC. It remains to show the measure conditions agree, and congruent angles are exactly those with equal measures.
If m\angle ABD = m\angle DBC = t, then m\angle ABC = 2t, so m\angle ABD = t = \tfrac{1}{2}\, m\angle ABC.
If m\angle ABD = \tfrac{1}{2}\, m\angle ABC, then
So the two conditions hold for exactly the same rays, and whichever is chosen as the definition, the other becomes a theorem. Without the interior clause the half condition is too loose. In Figure 9, m\angle ABC = 80^\circ and \overrightarrow{BE} makes m\angle ABE = 40^\circ, half of 80^\circ, but E is outside the angle, and \overrightarrow{BE} does not bisect it.
Boundaries
Only the hypothesis and conclusion swap. When the hypothesis has several parts, “the converse” may swap all of it or only one part, and those are different statements; say which.
A converse can depend on the geometry. The converse of the parallel-line theorems holds without the parallel postulate, so it survives in the geometry of the disc model in The parallel postulate, where the theorem itself fails. On a sphere, whose “lines” are great circles, even the converse fails: two meridians cross the equator at 90^\circ and still meet at the poles.
Definitions are different. A definition works in both directions by agreement, so its converse is true without proof (Why definitions work both ways, above). A theorem’s converse gets no such free pass, and neither does a postulate’s.
The contrapositive is not the converse. A theorem’s contrapositive is equivalent to it, so proving the contrapositive never settles the converse (Example 4).
Which statements are postulates is a choice. Euclid took “all right angles are equal” as a postulate; a course that measures angles proves it from two definitions. The same fact can be a postulate in one system and a theorem in another, so a reason is a postulate only relative to the list the course starts from.
Not every true biconditional is a definition. “A triangle has two congruent sides if and only if it has two congruent angles” is true both ways, but it links two ideas each defined already, and each direction needed its own proof: the base angles theorem and its converse. It is a pair of theorems.
A test needs something to test against. A counterexample to a proposed definition comes from agreed examples of the idea or a definition already in use. A brand-new word has nothing to be wrong about: any condition defines something, if perhaps nothing useful.
The setting matters for definitions too. In plane geometry, where all lines lie in one plane, “lines that do not intersect” does define parallel lines. Example 7’s clause is needed only once lines can leave the plane.
A definition may not lean on itself. Each definition uses earlier words, and the chain stops at the undefined terms: point, line, plane. A definition that uses its own word, directly or round a loop, fixes nothing.
Where this goes next
Every reason column in a two-column proof leans on definitions run in both directions, as in Examples 1 and 2, and on theorems run only forwards. The true converses are geometry’s tests. Proving lines parallel runs the angle-pair theorems backwards; Converse of the Pythagorean theorem tests for a right angle; Converse of the base angles theorem for an isosceles triangle; converses of the parallelogram properties for a parallelogram, found the way Example 11 found its repair. Converse of the triangle proportionality theorem is the same move for similar triangles. Showing that two conditions are equivalent, as in Example 12, is an instance of proving an if and only if, which makes proving both directions a standard method, and a definition’s condition is always necessary and sufficient for the word it defines.
Exercises
Definition, postulate or theorem
- Say whether each fact is a definition, a postulate or a theorem in this lesson’s sense. (a) A right angle is an angle whose measure is 90^\circ. (b) Through any two points there is exactly one line. (c) Vertical angles are congruent. (d) If B is between A and C, then AB + BC = AC.
- A proof contains the line “\angle P and \angle Q are right angles, so \angle P \cong \angle Q.” Show how this step follows from definitions alone, and say what kind of reason “all right angles are congruent” is.
- In a figure, \overline{AB} and \overline{CD} look the same length, \angle ABC looks like a right angle, and B is drawn on a line between A and E. Which of these may a proof take from the figure, and which must be given?
Writing both directions of a definition
- Write the definition of congruent segments as a biconditional, then as two conditionals.
- Write the definition of the midpoint of a segment as two conditionals, and say which one you use to conclude that a point is a midpoint.
- “An angle is acute if its measure is between 0^\circ and 90^\circ.” You know m\angle P = 41^\circ. Which direction lets you conclude that \angle P is acute?
Writing the converse
Write the converse of each theorem, keeping its setting, and say whether the converse is true. If it is false, give a counterexample.
- If two parallel lines are cut by a transversal, then each pair of same side interior angles is supplementary.
- If a point lies on the perpendicular bisector of \overline{AB}, then it is equidistant from A and B.
- If a triangle is equilateral, then each of its angles measures 60^\circ.
Both directions of a statement
For each statement, decide whether the statement and its converse are true.
- If a quadrilateral is a square, then it is a rhombus.
- If a quadrilateral is a parallelogram, then both pairs of its opposite angles are congruent.
- If a triangle has sides 3, 4 and 5, then it is a right triangle.
- Combine the statement in Exercise 7 and its converse into one “if and only if” sentence.
Using a definition in either direction
- M is the midpoint of \overline{PQ}, PM = 4y - 3 and MQ = 2y + 9. Find y and PQ, and say which direction of the definition you used.
- \overrightarrow{BD} bisects \angle ABC, m\angle ABD = (2x + 11)^\circ and m\angle DBC = (4x - 7)^\circ. Find x and m\angle ABC.
- R, S and T lie on a line with S between R and T. RS = 3k - 2, ST = k + 10 and RT = 32. Is S the midpoint of \overline{RT}? Say which direction of the definition decides it.
- Two lines cross, and two neighboring angles at the crossing measure (3n + 6)^\circ and (5n - 10)^\circ. Find both angles, and decide whether the lines are perpendicular.
- On a number line, A is at -4 and B is at 10. A line crosses \overline{AB} at the point 3. Does the line bisect \overline{AB}?
Using the converses
- A triangle has sides 9, 40 and 41. Is it a right triangle? If so, which angle is right?
- A triangle has sides 7, 8 and 11. Is it a right triangle?
- A triangle has sides \displaystyle \frac{5}{2}, 6 and \displaystyle \frac{13}{2}. Is it a right triangle?
- Two lines are cut by a transversal. A pair of corresponding angles measure (3x + 15)^\circ and (5x - 25)^\circ. For which x are the lines parallel, and what is each angle then?
- Two lines are cut by a transversal. A pair of same side interior angles measure (3x + 14)^\circ and (5x - 2)^\circ. For which x are the lines parallel, and what are the angles then?
- In \triangle ABC, m\angle B = (4x + 6)^\circ and m\angle C = (6x - 14)^\circ. Find the value of x for which AB = AC, and the three angles of the triangle for that x.
Testing a proposed definition
For each proposal, test both directions. If it fails, give a counterexample, say whether it is too loose, too tight or both, and repair it.
- “An angle is a right angle if it is not acute.”
- “The midpoint of \overline{AB} is a point that lies on \overline{AB}.”
- “Two angles are congruent if they are vertical angles.”
- “Two lines are perpendicular if they intersect.”
- “An angle is obtuse if its measure is greater than 90^\circ.” (Angle measures here run up to 180^\circ.)
- “A ray bisects \angle ABC if it starts at B and passes through the interior of \angle ABC.”
- “An angle is acute if its measure is less than the measure of its supplement.” (Take angles with measures strictly between 0^\circ and 180^\circ.)
Theorem or definition
- Every linear pair is supplementary. Explain, with a counterexample, why “two angles form a linear pair if they are supplementary” would not work as a definition.
- “Two lines cut by a transversal are parallel if and only if their corresponding angles are congruent” is true in both directions. Why is it called a theorem and not a definition?
Reasoning
- Two angles each measure 35^\circ. Can you conclude that they are vertical angles? Explain.
- Explain why a counterexample to the converse of a true theorem must satisfy the theorem’s conclusion but not its hypothesis.
- A student writes: “\angle 3 and \angle 6 are alternate interior angles and \angle 3 \cong \angle 6, so l \parallel m by the Alternate Interior Angles Theorem.” What is wrong with the reason, and what should it say?
- The proof of the converse of the Pythagorean theorem in this lesson uses the Pythagorean theorem. Explain why that is not circular.
- Give a theorem about quadrilaterals, not one from this lesson, whose converse is false, with a counterexample.
- A proof contains the line “AM = MB, so M is the midpoint of \overline{AB} (definition of midpoint).” What is missing? Give coordinates for points that show why it matters.
- Explain why a definition can be neither proved nor disproved, and yet a proposed definition can still be shown to be a bad one.
- A book defines perpendicular lines as lines that form a right angle. Suppose it then defined a right angle as an angle formed by perpendicular lines. Explain why neither word would have a meaning.
Answers
- (a) Definition: it gives “right angle” its meaning. (b) Postulate: it is accepted without proof. (c) Theorem: the Vertical Angles Theorem, proved from linear pairs. (d) Postulate: the segment addition postulate.
- Each angle measures 90^\circ (definition of right angle, forward), and angles with equal measures are congruent (definition of congruent angles, backward). So “all right angles are congruent” is a theorem here, proved from two definitions. Euclid, who had no angle measure, took it as his fourth postulate.
- Only that B lies between A and E on one line: a figure may be trusted for which points lie on which lines and in what order. The equal lengths and the right angle must be given, or marked with ticks and a square, or proved.
- \overline{AB} \cong \overline{CD} \iff AB = CD. Forward: congruent segments have the same length. Backward: segments with the same length are congruent.
- Forward: if M is the midpoint of \overline{AB}, then M lies on \overline{AB} and AM = MB. Backward: if M lies on \overline{AB} and AM = MB, then M is the midpoint. The backward one concludes it.
- The backward direction: 41^\circ is between 0^\circ and 90^\circ, so the angle is acute.
- Converse: if two lines are cut by a transversal and a pair of same side interior angles is supplementary, then the lines are parallel. True; it is one of the parallel-line converses.
- Converse: if a point is equidistant from A and B, then it lies on the perpendicular bisector of \overline{AB}. True, and the proof waits for a later lesson. Let M be the midpoint, which is on the bisector. A point P \ne M equidistant from A and B is off line AB, and triangles PMA and PMB match side for side. SSS, one of the congruence criteria set out later, will make them congruent, so that \angle PMA \cong \angle PMB; they form a linear pair, so each is 90^\circ and \overleftrightarrow{PM} is the perpendicular bisector. The same argument is Exercise 12 of proving the Base Angles Theorem.
- Converse: if each angle of a triangle measures 60^\circ, then it is equilateral. True. \angle B \cong \angle C gives AB = AC and \angle A \cong \angle B gives BC = AC, by the converse of the Base Angles Theorem, planned in Example 10 and proved in a later lesson; the result is Corollary 2 of equilateral and equiangular.
- The statement is true: a square has four equal sides. The converse, “a rhombus is a square”, is false: a rhombus with angles 60^\circ and 120^\circ is not a square.
- Both true. For the converse, call the angles x, y, x, y in order around ABCD. They add to 360^\circ, so x + y = 180^\circ. Then \angle A and \angle B are supplementary same side interior angles, so \overline{AD} \parallel \overline{BC}; likewise \angle A and \angle D give \overline{AB} \parallel \overline{DC}.
- The statement is true by the converse of the Pythagorean theorem, since 3^2 + 4^2 = 25 = 5^2. The converse, “a right triangle has sides 3, 4 and 5”, is false: 5^2 + 12^2 = 169 = 13^2 gives a right triangle with sides 5, 12 and 13.
- Two lines cut by a transversal are parallel if and only if a pair of same side interior angles is supplementary.
- y = 6 and PQ = 42. Forward: 4y - 3 = 2y + 9, so 2y = 12. Each half is 4(6) - 3 = 21.
- x = 9 and m\angle ABC = 58^\circ. Forward: 2x + 11 = 4x - 7, so 2x = 18. Each half is 2(9) + 11 = 29^\circ.
- Yes. (3k - 2) + (k + 10) = 32 gives 4k = 24, so k = 6, RS = 16 and ST = 16. S is on \overline{RT} with RS = ST, so the backward direction makes it the midpoint.
- 75^\circ and 105^\circ; not perpendicular. A linear pair: (3n + 6) + (5n - 10) = 180, so n = 23. The four angles are 75^\circ, 105^\circ, 75^\circ, 105^\circ, none right, and perpendicular lines form a right angle (forward), so these lines are not perpendicular.
- Yes. 3 - (-4) = 7 and 10 - 3 = 7, so the point is the midpoint (backward), and a line through the midpoint bisects the segment (backward again).
- Yes. 9^2 + 40^2 = 81 + 1600 = 1681 = 41^2, so by the converse the angle opposite the side of 41 is right.
- No. 7^2 + 8^2 = 113 but 11^2 = 121. The only possible right angle is opposite the longest side, and the theorem rules it out.
- Yes. \displaystyle \left(\frac{5}{2}\right)^2 + 6^2 = \frac{25}{4} + 36 = \frac{169}{4} = \left(\frac{13}{2}\right)^2, so the angle opposite the longest side is right.
- x = 20, and each angle is 75^\circ. By the converse, parallel when 3x + 15 = 5x - 25, so 2x = 40; and by the theorem, no other x works.
- x = 21, and the angles are 77^\circ and 103^\circ. Parallel when the pair is supplementary: (3x + 14) + (5x - 2) = 180, so 8x = 168.
- x = 10; the angles are m\angle B = 46^\circ, m\angle C = 46^\circ, m\angle A = 88^\circ. AB = AC exactly when \angle B \cong \angle C: 4x + 6 = 6x - 14 gives x = 10, and 180^\circ - 2(46^\circ) = 88^\circ.
- Backward fails: 120^\circ is not acute and not right. Too loose. Repair: its measure is 90^\circ.
- Backward fails: a point of \overline{AB} very near A is not the midpoint. Too loose. Repair: add AM = MB.
- Forward fails: two 50^\circ angles in different figures are congruent and not vertical. Too tight. (Backward holds, by the Vertical Angles Theorem.) Repair: same measure.
- Backward fails: lines crossing at 40^\circ intersect and are not perpendicular. Too loose. Repair: add “to form a right angle”.
- Backward fails: a straight angle, 180^\circ, is over 90^\circ and not obtuse. Too loose. Repair: between 90^\circ and 180^\circ.
- Backward fails: in a 70^\circ angle, a ray making 20^\circ with one side and 50^\circ with the other passes through the interior and does not bisect. Too loose. Repair: add that the two angles it makes are congruent.
- It works both ways. The supplement measures 180^\circ - m, and m < 180 - m exactly when 2m < 180, that is, m < 90. The boundary m = 180 - m gives m = 90, a right angle.
- A 130^\circ angle and a 50^\circ angle drawn apart are supplementary but share no side, so they are not a linear pair. The converse of a true theorem fails, so it cannot define.
- “Parallel” and “corresponding angles” were each defined already. The statement links two meanings that existed before it, so each direction had to be proved; nothing in it was agreed.
- No. That would use the converse of the Vertical Angles Theorem, which is false (Example 3). Only their position, opposite each other at one crossing, makes angles vertical.
- The converse reads “if conclusion, then hypothesis”. A counterexample to it is a case where its “if” part holds and its “then” part fails: the original conclusion true and the original hypothesis false.
- That theorem assumes l \parallel m and concludes the angles are congruent. Here the angles are given and parallel lines are the goal, so the reason must be the converse of the Alternate Interior Angles Theorem.
- The theorem was proved first, on its own, and is applied to a new right triangle DEF that we built. Nothing assumes the converse, which is the claim about \triangle ABC.
- For example: if a quadrilateral is a rhombus, then its diagonals are perpendicular. The converse fails for the kite with vertices (-1, 0), (0, 2), (1, 0), (0, -1): its diagonals lie on the axes, so they are perpendicular, but its sides are \displaystyle \sqrt{5}, \displaystyle \sqrt{5}, \displaystyle \sqrt{2}, \displaystyle \sqrt{2}, so it is not a rhombus.
- It never says M lies on \overline{AB}. With A = (0, 0), B = (6, 0) and M = (3, 4), AM = MB = 5, yet M is off the segment.
- A definition creates the meaning, so there is nothing earlier for it to be true or false about. A proposal meant to capture an idea already fixed by agreed examples can miss it, and one counterexample in either direction shows that.
- To tell whether an angle is right you would ask whether its lines are perpendicular, which would send you back to whether the angle is right. Nothing outside the loop fixes either word. A right angle is defined instead by its measure, 90^\circ.