Lesson 1.5 · Unit 1 · G-CO.1
Writing a definition in the form a reason column can use
A definition in geometry is not a description; it is a tool. Every definition works in both directions, and a proof uses it one way to conclude something and the other way to establish a condition. Learning to state a definition in that usable form is most of what makes early proofs go smoothly.
- Two lines are perpendicular when they intersect to form a right angle, written \( \overleftrightarrow{AB} \perp \overleftrightarrow{CD} \).
- One right angle is enough. By the theorem at the end of lesson 1.4, if one of the four angles is right then all four are, so a proof need only establish one.
- A segment bisector is any line, ray or segment through the midpoint, so it divides the segment into two congruent segments.
- A perpendicular bisector does both: it passes through the midpoint and is perpendicular to the segment. Both conditions must be established to claim it.
- An angle bisector is a ray through the interior dividing an angle into two congruent angles.
- Every definition is a biconditional, so it works in both directions. If a ray bisects an angle then the parts are congruent, and if the parts are congruent then the ray bisects.
- A theorem usually works in only one direction, which is the difference that makes definitions so useful in proofs and makes lesson 2.2 necessary.
- Cite the definition by name in the reason column, as "definition of angle bisector" rather than "because it splits it evenly."
Where students lose marks: claiming a perpendicular bisector from one condition. A line through the midpoint is a bisector; a line perpendicular to the segment is a perpendicular line. Only a line that is both is the perpendicular bisector, and a proof must establish both.
The problem. \( A \) is \( (2, 3) \) and \( B \) is \( (8, 11) \). (a) Find the equation of the perpendicular bisector of \( \overline{AB} \). (b) Verify that the point \( (9, 4) \) lies on it, and check that this point is equidistant from \( A \) and \( B \). (c) State the definition of perpendicular bisector in the two directions a proof uses.
Step one: identify what the definition requires. A perpendicular bisector must pass through the midpoint and be perpendicular to the segment. So two things are needed: a point and a slope.
Step two: find the midpoint. \( M = \left( \dfrac{2 + 8}{2}, \dfrac{3 + 11}{2} \right) = (5, 7) \). Both coordinates lie between those of \( A \) and \( B \), as a midpoint's must.
Step three: find the slope of \( \overline{AB} \). \( m = \dfrac{11 - 3}{8 - 2} = \dfrac{8}{6} = \dfrac{4}{3} \).
Step four: find the perpendicular slope. Perpendicular slopes are negative reciprocals, so the bisector has slope \( -\dfrac{3}{4} \). Check: \( \dfrac{4}{3} \times \left(-\dfrac{3}{4}\right) = -1 \), as perpendicular slopes must multiply to.
Step five: write the equation and confirm both conditions. Using point-slope form with the midpoint: \[ y - 7 = -\frac{3}{4}(x - 5) \] It passes through \( (5,7) \), the midpoint, satisfying the bisector condition, and its slope is the negative reciprocal of the segment's, satisfying the perpendicular condition. Both halves of the definition are met, which is what licenses the name.
Step six: verify \( (9, 4) \) lies on the line. Substituting: \( 4 - 7 = -3 \) on the left, and \( -\dfrac{3}{4}(9 - 5) = -\dfrac{3}{4}(4) = -3 \) on the right. They agree, so the point is on the perpendicular bisector.
Step seven: check the distances. From \( (9,4) \) to \( A(2,3) \): differences 7 and 1, so \( \sqrt{49 + 1} = \sqrt{50} \). From \( (9,4) \) to \( B(8,11) \): differences 1 and 7, so \( \sqrt{1 + 49} = \sqrt{50} \). Equal, so the point is equidistant from \( A \) and \( B \). This is not a coincidence; it is the perpendicular bisector theorem, proved in lesson 6.1.
Step eight: state the definition both ways. Forward: if \( \ell \) is the perpendicular bisector of \( \overline{AB} \), then \( \ell \) passes through the midpoint of \( \overline{AB} \) and \( \ell \perp \overline{AB} \). This is the direction used to extract facts from a given. Backward: if a line passes through the midpoint of \( \overline{AB} \) and is perpendicular to \( \overline{AB} \), then it is the perpendicular bisector. This is the direction used to conclude that a line deserves the name. Both directions are valid because this is a definition. A theorem would need its converse proved separately before it could be used backward, which is exactly the trap lesson 3.3 covers.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- What does \( \perp \) mean?
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Perpendicular: the lines meet at a right angle
- Two perpendicular lines meet. How many right angles are formed?
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Four
- A line passes through the midpoint of a segment. Is it the perpendicular bisector?
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Only one condition is met. Not necessarily; it must also be perpendicular
- Find the slope perpendicular to a line of slope \( \frac{2}{5} \).
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\( -\frac{5}{2} \)
- What reason justifies writing \( \angle ABD \cong \angle DBC \) when
\( \overrightarrow{BD} \) bisects \( \angle ABC \)?
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The definition of an angle bisector
- Find the perpendicular bisector of the segment from \( (1, 2) \) to \( (7, 6) \).
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Midpoint: \( \left( \dfrac{1+7}{2}, \dfrac{2+6}{2} \right) = (4, 4) \). Slope of the segment: \( \dfrac{6-2}{7-1} = \dfrac{4}{6} = \dfrac{2}{3} \). Perpendicular slope: \( -\dfrac{3}{2} \). Equation: \( y - 4 = -\dfrac{3}{2}(x - 4) \). Check that it passes through the midpoint: at \( x = 4 \) the right side is 0, so \( y = 4 \). Correct. \( y - 4 = -\frac{3}{2}(x - 4) \)
- Verify that \( (7, 4) \) is equidistant from \( (1, 2) \) and \( (7, 6) \), and say what
that suggests.
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To \( (1,2) \): differences 6 and 2, so \( \sqrt{36 + 4} = \sqrt{40} \). To \( (7,6) \): differences 0 and 2, so \( \sqrt{0 + 4} = 2 \). These are not equal, since \( \sqrt{40} \approx 6.3 \). So \( (7,4) \) is not equidistant. Checking against the previous problem's answer: substituting \( x = 7 \) into \( y - 4 = -\frac{3}{2}(x - 4) \) gives \( y - 4 = -4.5 \), so \( y = -0.5 \), not 4. The point is not on the perpendicular bisector, which is consistent with the distances being unequal. The two facts always agree, which is the content of the perpendicular bisector theorem: a point is equidistant from the endpoints exactly when it lies on the perpendicular bisector. It is not equidistant, and correspondingly it does not lie on the perpendicular bisector
- Explain why a definition can be used in both directions in a proof but a theorem usually
cannot.
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A definition states what a word means, so it is a biconditional: the term applies exactly when the condition holds. "A midpoint is a point that divides a segment into two congruent segments" means both that every midpoint does this and that anything doing this is a midpoint. A proof may therefore use it either way, citing "definition of midpoint" in each case. A theorem is a one-directional claim: if this, then that. Proving it establishes only that direction. "If two angles form a linear pair, then they are supplementary" does not by itself license concluding a linear pair from supplementarity, and in that case the converse is genuinely false, since supplementary angles need not be adjacent. A converse has to be proved separately, and some converses are true while others are not. Treating a theorem as reversible is one of the four errors this course names, and unit 3 is largely about keeping each theorem apart from its converse. A definition is a biconditional; a theorem states one direction, and its converse needs its own proof
- \( \overrightarrow{BD} \) bisects \( \angle ABC \). \( m\angle ABD = (4x - 7)^\circ \)
and \( m\angle DBC = (2x + 9)^\circ \). Is \( \angle ABC \) acute, right or obtuse?
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By the definition of an angle bisector the two parts are congruent, so their measures are equal: \( 4x - 7 = 2x + 9 \), giving \( 2x = 16 \) and \( x = 8 \). Each part measures \( 4(8) - 7 = 25^\circ \), confirmed by the other expression: \( 2(8) + 9 = 25^\circ \). By the angle addition postulate, \( m\angle ABC = 25 + 25 = 50^\circ \). Since \( 50 \lt 90 \), the angle is acute. Acute, measuring \( 50^\circ \)
- A line \( \ell \) is perpendicular to \( \overline{AB} \) and intersects it at a point
\( P \) with \( AP = 5 \) and \( PB = 5 \). Prove \( \ell \) is the perpendicular bisector,
naming every reason.
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Given: \( \ell \perp \overline{AB} \), \( \ell \) meets \( \overline{AB} \) at \( P \), \( AP = 5 \), \( PB = 5 \). Prove: \( \ell \) is the perpendicular bisector of \( \overline{AB} \). 1. \( AP = 5 \) and \( PB = 5 \). Reason: given. 2. \( AP = PB \). Reason: transitive property of equality, since both equal 5. 3. \( \overline{AP} \cong \overline{PB} \). Reason: definition of congruent segments. 4. \( P \) lies on \( \overline{AB} \) between \( A \) and \( B \). Reason: given, since \( \ell \) intersects the segment at \( P \). 5. \( P \) is the midpoint of \( \overline{AB} \). Reason: definition of midpoint, using steps 3 and 4. 6. \( \ell \) passes through the midpoint of \( \overline{AB} \). Reason: step 5 and the given that \( \ell \) meets the segment at \( P \). 7. \( \ell \perp \overline{AB} \). Reason: given. 8. \( \ell \) is the perpendicular bisector of \( \overline{AB} \). Reason: definition of perpendicular bisector, using steps 6 and 7. Both halves of the definition were established separately and then combined, which is exactly what the definition demands. Step 4 is the one students omit: without betweenness, a point equidistant from \( A \) and \( B \) need not be on the segment at all. Proved, with both conditions of the definition established