HSG.CO.C.9: Proving Theorems About Lines and Angles
In plain English: HSG.CO.C.9 is the Common Core geometry standard that asks students to prove theorems about lines and angles, not just use them. Students prove that vertical angles are congruent, that a transversal of parallel lines forms congruent alternate interior and corresponding angles, and that the perpendicular bisector of a segment is exactly the set of points equidistant from its endpoints, in high school Geometry.
Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints.
Common Core State Standards for Mathematics · Domain: Congruence (CO) · Cluster: Prove geometric theorems Also written as HSG-CO.C.9 or G-CO.9 · Official standard
Students move from using angle facts to proving them. Starting from the linear pair postulate, rigid motions and the parallel postulate, they prove four theorems named in HSG.CO.C.9: vertical angles are congruent; when a transversal crosses parallel lines, corresponding angles are congruent and alternate interior angles are congruent; and the perpendicular bisector of a segment is exactly the set of points equidistant from the endpoints. Each proof is written as a two-column proof or a short paragraph in which every statement has a reason.
The lesson shows how theorems build on one another: the vertical angles theorem and the corresponding angles theorem together prove the alternate interior angles theorem. For the perpendicular bisector, students prove both directions, because "exactly those" points means the statement and its converse.
Learning Objectives
By the end of this lesson, students will be able to:
Prove that vertical angles are congruent using the linear pair postulate
Prove that corresponding angles formed by a transversal of parallel lines are congruent, using a translation and the parallel postulate
Prove that alternate interior angles are congruent by chaining the corresponding angles and vertical angles theorems
Prove that a point on the perpendicular bisector of a segment is equidistant from its endpoints, and prove the converse
Use these theorems to find unknown angle measures and to decide whether a point lies on a perpendicular bisector
Prior Knowledge Required
Students should already be comfortable with:
Supplementary, complementary, vertical and adjacent angles, used to write and solve equations 7.G.B.5
Informal arguments about the angles formed when parallel lines are cut by a transversal 8.G.A.5
Precise definitions of angle, perpendicular lines, parallel lines and line segment HSG.CO.A.1
Triangle congruence by SAS and SSS HSG.CO.B.8
The distance formula in the coordinate plane 8.G.B.8
Open a pair of scissors and hold it up, or sketch two crossing lines on the board. Label the angle between the blades 38°.
Warm-Up Prompt
"The blades of the scissors make a 38° angle. Find the other three angles formed by the two blades. Then explain how you know, without measuring, that the angle across from the 38° angle is also 38°."
Students find 142°, 38° and 142°. Collect explanations. Most will use "180 minus 38" twice. Point out that this reasoning works for any angle, not only 38°, so it can become a proof. Tell students that today every angle fact from middle school gets a reason, and that a proof must work for every figure, not just the one drawn.
Direct Instruction20 minutes
What we may assume. Write the starting facts on the board. A proof can only use definitions, these assumptions and theorems already proved.
Linear pair postulate: if two angles form a linear pair, they are supplementary (their measures add to 180°).
Parallel postulate: through a point not on a line there is exactly one line parallel to the given line.
Rigid motions: translations, rotations and reflections preserve distances and angle measures, and a translation maps a line to a parallel line (or to itself).
Triangle congruence: SAS and SSS, from HSG.CO.B.8.
Theorem 1, vertical angles. Two lines meet, forming ∠1, ∠2 and ∠3 in order around the point, so ∠1 and ∠3 are vertical. ∠1 and ∠2 form a linear pair, and so do ∠2 and ∠3. So m∠1 + m∠2 = 180° and m∠2 + m∠3 = 180°. Subtract m∠2 from both: m∠1 = m∠3, so ∠1 ≅ ∠3.
Theorem 2, corresponding angles. Use Diagram 1. Translate the plane along t so that P moves to Q. The translation maps t onto itself and maps m to a line through Q parallel to m. By the parallel postulate, that line is n. So ∠2 lands on ∠6, and because a translation preserves angle measure, ∠2 ≅ ∠6. (Some textbooks take this statement as a postulate instead. Tell students which one your course uses.)
Theorem 3, alternate interior angles. ∠3 ≅ ∠2 (vertical angles) and ∠2 ≅ ∠6 (corresponding angles), so ∠3 ≅ ∠6 by the transitive property. The first two theorems prove the third.
Theorem 4, perpendicular bisector. Let ℓ be the perpendicular bisector of AB at M. If P is on ℓ and not at M, then AM ≅ BM, ∠PMA ≅ ∠PMB (both right angles) and PM ≅ PM, so △PMA ≅ △PMB by SAS and PA = PB. (If P is M, then PA = PB because M is the midpoint.) Conversely, if PA = PB and P is not on line AB, then △PMA ≅ △PMB by SSS, so ∠PMA ≅ ∠PMB. They form a linear pair, so each is 90°, and line PM is ℓ. (If P is on line AB, then PA = PB makes P the midpoint M.) Both directions together say the points of ℓ are exactly the points equidistant from A and B. Show the coordinate check in Diagram 2.
Vertical angles, with algebra
Two lines intersect. One angle measures (3x + 10)° and the angle vertical to it measures (5x - 30)°. Find x and all four angles.
Equation: Vertical angles are congruent: 3x + 10 = 5x - 30, so x = 20. The pair measures 70°, and each of the other two angles is 180° - 70° = 110°
Corresponding angles
In Diagram 1, m ∥ n, m∠2 = (2x + 15)° and m∠6 = (4x - 35)°. Find m∠2.
Equation: Corresponding angles theorem: 2x + 15 = 4x - 35, so x = 25 and m∠2 = m∠6 = 65°
Alternate interior angles
With the angle numbering of Diagram 1, m ∥ n, m∠4 = (7x - 12)° and m∠5 = (4x + 27)°. Find x and m∠4, and give the chain of theorems.
Equation: ∠4 ≅ ∠1 (vertical) and ∠1 ≅ ∠5 (corresponding), so ∠4 ≅ ∠5: 7x - 12 = 4x + 27, x = 13, m∠4 = 79°
Perpendicular bisector theorem
Segment AB has A(1, 1) and B(7, 5). Point P(2, 6) lies on the perpendicular bisector of AB. Confirm that P is equidistant from A and B.
Equation: PA = √((2 - 1)² + (6 - 1)²) = √26 and PB = √((2 - 7)² + (6 - 5)²) = √26, so PA = PB
Converse of the perpendicular bisector theorem
Find the point Q on the x-axis that is equidistant from A(1, 1) and B(7, 5). Then explain why Q must lie on the perpendicular bisector of AB.
Equation: (x - 1)² + 1 = (x - 7)² + 25 gives 12x = 72, so Q = (6, 0). QA = QB, so by the converse Q is on the perpendicular bisector (Q - M = (2, -3) is perpendicular to AB)
After Example 3, ask: "Which step would fail if m and n were not parallel?" (Step 2, the corresponding angles theorem.) After Example 5, stress that the converse is what lets us conclude a point is on ℓ from distances alone.
Guided Practice15 minutes
Pairs complete two proofs. First, they prove the second pair of alternate interior angles congruent. Give the statements with the reasons blank, then reveal the completed proof below.
Two-column proof that angle 4 is congruent to angle 5
Statement
Reason
1. m ∥ n, and t is a transversal
Given
2. ∠1 ≅ ∠5
Corresponding angles theorem (m ∥ n)
3. ∠1 ≅ ∠4
Vertical angles theorem
4. ∠4 ≅ ∠5
Transitive property of congruence (steps 2 and 3)
Second, pairs write a paragraph proof of the perpendicular bisector theorem using a reflection instead of SAS: the reflection across ℓ maps A to B (ℓ is the perpendicular bisector of AB) and leaves P fixed (P is on ℓ), so it maps segment PA onto segment PB, and reflections preserve distance. Circulate and ask each pair: "What does each step of your proof depend on?" Listen for students who write "it looks congruent" and ask them for the theorem instead.
Independent Practice15 minutes
Students work alone and justify every answer with a theorem. (1) Two lines intersect, forming vertical angles of (6x - 8)° and (4x + 14)°. Find x and all four angles. (x = 11: the vertical pair is 58° and the other two angles are 122°.) (2) With the angle numbering of Diagram 1 and m ∥ n, m∠5 = 104°. Find m∠1, m∠4, m∠8 and m∠6. (∠1 = 104° by corresponding angles, ∠4 = 104° by vertical angles with ∠1, ∠8 = 104° by vertical angles with ∠5, and ∠6 = 76° by the linear pair postulate.) (3) Is R(9, 1) on the perpendicular bisector of the segment from (1, 1) to (7, 5)? (No: its distances are 8 and √20, which are not equal, so by the theorem R cannot be on it.)
Closure5-10 minutes
Exit ticket: (1) With the angle numbering of Diagram 1 and m ∥ n, m∠7 = 131°. Name three other angles that measure 131° and give the theorem for each. (∠6 by vertical angles, ∠3 by corresponding angles, ∠2 by vertical angles with ∠3.) (2) Write the perpendicular bisector theorem and its converse as two separate if-then statements. (3) Which theorem did you need to prove the alternate interior angles theorem?
Differentiation Strategies
For Struggling Students
Give a reason bank (Given, Linear pair postulate, Vertical angles theorem, Corresponding angles theorem, Transitive property, SAS, SSS, Definition of midpoint) so students can focus on the order of the argument
Have students color-code each pair of angles on Diagram 1 before writing any proof
Start proofs with a numeric case (for example 38°) and then replace the number with m∠1
For Advanced Students
Prove the converse of the corresponding angles theorem: if corresponding angles are congruent, the lines are parallel (hint: suppose they meet and find a contradiction with the triangle angle sum or the parallel postulate)
Prove the alternate interior angles theorem directly with a 180° rotation about the midpoint of segment PQ
Explain why the perpendicular bisector theorem needs both directions to describe the set of all points equidistant from A and B
Assessment Guidance
What to Look For
Strong proofs give a reason for every statement, and the reason is a definition, a postulate or a theorem proved earlier, never "it looks like it". Check that students do not use the theorem they are proving as a reason. For parallel-line theorems, students should state that the lines are parallel before using corresponding angles. For the perpendicular bisector, look for both directions: a point on the bisector is equidistant, and an equidistant point is on the bisector.
02
Classroom Activities
3 Activities
1
Proof Strip Sort
20 minGroups of 3
Each group receives two envelopes with 20 strips in all. Envelope A holds the 8 strips of the vertical angles proof and Envelope B holds the 12 strips of the converse of the perpendicular bisector theorem. Groups rebuild each proof in two columns, matching every statement to its reason.
Envelope A: Vertical Angles (8 strips)
Statements: "∠1 and ∠2 form a linear pair", "∠2 and ∠3 form a linear pair", "m∠1 + m∠2 = 180° and m∠2 + m∠3 = 180°", "m∠1 = m∠3, so ∠1 ≅ ∠3"
Reasons: "Two lines intersect (given), so the angles share a side and their other sides are opposite rays", "Same reason, for ∠2 and ∠3", "Linear pair postulate", "Subtract m∠2 from both equations (subtraction property of equality)"
Envelope B: Converse of the Perpendicular Bisector Theorem (12 strips)
Statements: "PA = PB, and M is the midpoint of AB", "AM = BM", "PM = PM", "△PMA ≅ △PMB", "∠PMA ≅ ∠PMB", "∠PMA and ∠PMB are right angles, so line PM is the perpendicular bisector of AB"
Reasons: "Given", "Definition of midpoint", "Reflexive property", "SSS", "Corresponding parts of congruent triangles are congruent", "Congruent angles that form a linear pair each measure 90°"
Procedure
Groups lay the statements in order first, then place the reasons next to them
Each group writes one sentence explaining why the vertical angles proof works for every pair of intersecting lines
Groups compare with a neighboring group and resolve any differences
Discussion Questions
Which strip in Envelope B depends on the linear pair postulate?
Could the strips of Envelope B be put in a different order and still form a valid proof?
2
Patty Paper Transversals
20 minPairs
Students use tracing paper to act out the rigid motions behind the parallel line theorems. A translation along the transversal shows corresponding angles are congruent, and a half-turn about the midpoint of the transversal segment shows alternate interior angles are congruent.
Procedure
Draw two parallel lines with a ruler (trace both edges of the ruler) and a transversal. Number the eight angles as in Diagram 1
Trace line m, the transversal and ∠2 on patty paper. Slide the patty paper along the transversal until the traced point lands on Q. Record which angle the traced ∠2 covers
Mark the midpoint of the transversal segment between the lines. Trace ∠3, put a pencil point on the midpoint and turn the patty paper 180°. Record which angle the traced ∠3 covers
Repeat on a second sheet where the two lines are not parallel, and describe what goes wrong
Discussion Questions
Where in the slide did you use the fact that the lines are parallel?
Why is the half-turn a proof that works for every pair of parallel lines, not just the ones you drew?
Modification for Distance Learning
Students build the figure in dynamic geometry software, use the translate-by-vector tool along the transversal and the rotate-by-180° tool about the midpoint, and screenshot the image angles on top of their partners.
3
Fold, Measure and Locate
15 minPairs
Students fold paper to make a perpendicular bisector, test points on it with string, and then use the converse to solve a location problem.
Procedure
Mark two points A and B about 12 cm apart on plain paper. Fold the paper so that A lands exactly on B and crease it. Explain why the crease is the perpendicular bisector of AB
Mark three points on the crease. For each one, stretch a string from the point to A, pinch it, and swing it to B. Record whether the lengths match
Mark three points off the crease and repeat
Location problem: A and B are two bus stops. Find every place on the paper where a bench is the same distance from both stops, and justify the answer with the converse
Challenge Variation
Add a third bus stop C that is not on line AB. Fold the perpendicular bisectors of AB and BC. Explain, with the theorem and its converse, why the point where the creases cross is the same distance from all three stops.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Parallel Lines Cut by a Transversal
Lines m and n are parallel and t is a transversal meeting them at P and Q. The angles are numbered 1-4 at P and 5-8 at Q (1 and 5 above-left, 2 and 6 above-right, 3 and 7 below-left, 4 and 8 below-right). The highlighted angles ∠2, ∠3 and ∠6 are congruent: ∠2 and ∠6 are corresponding, ∠2 and ∠3 are vertical, and ∠3 and ∠6 are alternate interior. The figure is drawn to scale with m∠2 = 65°.
Diagram 2: The Perpendicular Bisector as a Set of Points
Line ℓ is the perpendicular bisector of AB, drawn to scale on a grid. P(2, 6) and Q(6, 0) are on ℓ, and each is √26 units from both A and B (double tick marks). The single tick marks show AM = MB.
04
Homework Assignment
~30 min
HSG.CO.C.9 Homework: Proving Line and Angle Theorems
Directions: Show all work. Give a reason (a definition, postulate or theorem) for every statement in a proof and for every angle or distance you find. Use the angle numbering of Diagram 1 when a problem refers to it.
Part 1: Vertical and Parallel-Line Angles (Problems 1-3)
Two lines intersect. One angle measures (8x - 20)° and the angle vertical to it measures (5x + 22)°. (a) Find x and all four angle measures. (b) Write a two-column proof that vertical angles are congruent, using ∠A, ∠B and ∠C for three consecutive angles around the intersection.
With the angle numbering of Diagram 1, m ∥ n, m∠2 = (3y + 16)° and m∠6 = (5y - 20)°. (a) Find y and m∠2. (b) Find m∠3, m∠7 and m∠5, naming the theorem or postulate you use for each.
Parallel lines p and q are cut by a transversal r, and r is perpendicular to p. Prove that r is also perpendicular to q. (Hint: use the corresponding angles theorem and the definition of perpendicular lines.)
Part 2: Perpendicular Bisectors (Problems 4-6)
Segment CD has endpoints C(-2, 1) and D(4, -3). (a) Find the midpoint of CD. (b) Show that E(3, 2) is equidistant from C and D. What does the converse of the perpendicular bisector theorem tell you about E? (c) Is F(0, 2) on the perpendicular bisector of CD? Justify with distances.
Two cell towers stand at points A and B. A repair crew wants a base camp the same distance from both towers. Write a two-column proof that any point P with PA = PB lies on the perpendicular bisector of AB. Then describe the set of all possible camp locations.
Find the point on the y-axis that is equidistant from G(-4, 2) and H(2, 6). Then show that this point lies on the line through the midpoint of GH perpendicular to GH, and explain which theorem guarantees this.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Logical Order
Every statement follows from earlier ones
One step out of order or skipped
Steps do not connect
Reasons
Every statement has a correct reason
One or two reasons missing or wrong
Reasons mostly missing
Computation
Angles and distances correct
Correct setup with an arithmetic error
Setup incorrect
Theorem and Converse
Uses the right direction of each theorem
Confuses direction once
Direction not considered
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Choose an answer for each multiple-choice question to see whether it is right and why. For the short answers, write your proof or solution on paper before opening the answer. Some questions use the angle numbering of Diagram 1. Reset quiz clears all answers.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Two lines intersect, and one of the angles measures 47°. What is the measure of the angle vertical to it?
Answer: B
Vertical angles are congruent, so the angle across from the 47° angle also measures 47°. Choice A, 133°, is the angle that forms a linear pair with it. Choice C treats the angles as complementary, and choice D doubles the angle.
Question 2 of 20 · Multiple Choice
In the proof that vertical angles ∠1 and ∠3 are congruent, which reason justifies m∠1 + m∠2 = 180°?
Answer: A
∠1 and ∠2 form a linear pair, so they are supplementary by the linear pair postulate. Choice B uses the theorem being proved as a reason, which is circular. Choice C needs parallel lines and a transversal, which this figure does not have. Choice D would only apply if the angles were right angles.
Question 3 of 20 · Multiple Choice
Two lines intersect. A pair of vertical angles measure (5x - 14)° and (3x + 22)°. What is the measure of each of these angles?
Answer: C
Vertical angles are congruent: 5x - 14 = 3x + 22, so x = 18 and each angle is 5(18) - 14 = 76°. Choice A is the value of x, not the angle. Choice B is the supplement of 76°. Choice D comes from setting the sum equal to 180°, which treats vertical angles as a linear pair.
Question 4 of 20 · Multiple Choice
Use the angle numbering of Diagram 1. Lines m and n are parallel and m∠2 = 118°. Which angle must also measure 118°?
Answer: C
∠2 and ∠6 are corresponding angles, so ∠6 ≅ ∠2 and m∠6 = 118°. Choice A, ∠1, forms a linear pair with ∠2, so it measures 62°. Choice B, ∠4, is vertical to ∠1, so it is also 62°. Choice D, ∠5, corresponds to ∠1 and measures 62°.
Question 5 of 20 · Multiple Choice
Two parallel lines are cut by a transversal. A pair of alternate interior angles measure (6x + 4)° and (8x - 26)°. What is the measure of each angle?
Answer: D
Alternate interior angles of parallel lines are congruent: 6x + 4 = 8x - 26, so x = 15 and each angle is 6(15) + 4 = 94°. Choice A is x. Choice B is the supplement of 94°, the error of treating the angles as supplementary. Choice C halves the angle.
Question 6 of 20 · Multiple Choice
Lines p and q are cut by a transversal, and a pair of alternate interior angles measure 70° and 72°. What can you conclude?
Answer: B
If p and q were parallel, the alternate interior angles theorem would force the angles to be congruent. They are not (70° is not 72°), so p and q cannot be parallel. This is the contrapositive of the theorem. Choice A ignores the 2° difference, and choice D misses that the theorem still gives information. Choice C is unrelated to these angles.
Question 7 of 20 · Multiple Choice
Which rigid motion is used to prove that corresponding angles are congruent when a transversal t crosses parallel lines m and n at P and Q?
Answer: A
The translation by the vector from P to Q maps t onto itself and maps m to the line through Q parallel to m, which is n by the parallel postulate. So each angle at P lands on its corresponding angle at Q. Choice B maps the angles above m to angles below m at the same point. Choices C and D do not move P to Q.
Question 8 of 20 · Multiple Choice
A proof says: "∠3 ≅ ∠2 by the vertical angles theorem, and ∠2 ≅ ∠6 by the corresponding angles theorem." Which reason completes the proof that ∠3 ≅ ∠6?
Answer: D
Two angles congruent to the same angle are congruent to each other: that is the transitive property. Choice A is a triangle congruence criterion and there are no triangles here. Choice B gives supplementary angles, not congruent ones. Choice C is about segments.
Question 9 of 20 · Multiple Choice
Point P lies on the perpendicular bisector of segment AB, and PA = 14 cm. What is PB?
Answer: B
By the perpendicular bisector theorem, every point on the perpendicular bisector is equidistant from A and B, so PB = 14 cm. Choice A halves PA, confusing P with the midpoint. Choice C doubles it. Choice D ignores the theorem.
Question 10 of 20 · Multiple Choice
Segment AB has endpoints A(0, 3) and B(8, -1). Which point lies on the perpendicular bisector of AB?
Answer: D
(6, 5) is √(6² + 2²) = √40 from A and √(2² + 6²) = √40 from B, so by the converse of the perpendicular bisector theorem it is on the bisector. (4, 5) is √20 from A but √52 from B. (8, 3) is 8 from A and 4 from B. (2, 0) is √13 from A and √37 from B.
Question 11 of 20 · Multiple Choice
To prove that a point P on the perpendicular bisector of AB (with P not on AB) is equidistant from A and B, you show △PMA ≅ △PMB, where M is the midpoint. Which criterion applies?
Answer: A
AM ≅ BM (midpoint), ∠PMA ≅ ∠PMB (both right angles) and PM ≅ PM, which is side-angle-side. Choice B needs PA ≅ PB, which is what we want to prove. Choices C and D are not valid congruence criteria.
Question 12 of 20 · Multiple Choice
To prove the converse (if PA = PB, then P is on the perpendicular bisector of AB), a student shows △PMA ≅ △PMB by SSS, where M is the midpoint. What must the student show next?
Answer: C
The congruent triangles give ∠PMA ≅ ∠PMB. Two congruent angles that form a linear pair each measure 90°, so PM is perpendicular to AB at its midpoint. Choice A is the given, not a conclusion. Choice B is false in general. Choice D contradicts the perpendicular conclusion.
Question 13 of 20 · Multiple Choice
Why does the corresponding angles theorem require the two lines to be parallel?
Answer: B
Without parallel lines, the translation along the transversal does not map one line onto the other, and corresponding angles can have different measures (try a 60° and a 75° angle). Choice A is false. Choice C is false: a transversal can cross any two lines. Choice D is false: vertical angles are congruent for any two intersecting lines.
Question 14 of 20 · Multiple Choice
A straight road crosses two parallel railroad tracks. The acute angle between the road and the first track is 57°. What is the acute angle between the road and the second track?
Answer: A
The road is a transversal of the parallel tracks, and the acute angles at the two crossings are corresponding angles, so they are congruent: 57°. Choice B is the obtuse angle at the crossing. Choice C treats the angles as complementary, and choice D doubles 57°.
Question 15 of 20 · Short Answer
Lines j and k intersect at X, forming ∠5, ∠6 and ∠7 in order around X, so that ∠5 and ∠7 are vertical angles. Write a two-column proof that ∠5 ≅ ∠7.
1. ∠5 and ∠6 form a linear pair; ∠6 and ∠7 form a linear pair (the lines meet at X). 2. m∠5 + m∠6 = 180° and m∠6 + m∠7 = 180° (linear pair postulate). 3. m∠5 + m∠6 = m∠6 + m∠7 (substitution, both equal 180°). 4. m∠5 = m∠7 (subtraction property). 5. ∠5 ≅ ∠7 (definition of congruent angles).
Question 16 of 20 · Short Answer
Parallel lines a and b are cut by transversal c at points R (on a) and S (on b). Angle w lies between the lines at R, on the left of c. Angle z lies between the lines at S, on the right of c. Prove that ∠w ≅ ∠z.
Let ∠v be the angle at R on the side of a away from b, on the right of c. ∠w and ∠v are vertical angles, so ∠w ≅ ∠v (vertical angles theorem). ∠v and ∠z are in matching positions (on the same side of their lines, facing the same way, and on the right of c), so they are corresponding angles, and ∠v ≅ ∠z because a ∥ b (corresponding angles theorem). So ∠w ≅ ∠z by the transitive property. These are alternate interior angles.
Question 17 of 20 · Short Answer
Two parallel lines are cut by a transversal. A pair of corresponding angles measure (4k + 9)° and (6k - 25)°. Find k and the angle measure. Then find the measure of an angle that forms a linear pair with one of them.
Corresponding angles are congruent: 4k + 9 = 6k - 25, so k = 17 and each angle is 4(17) + 9 = 77°. An angle forming a linear pair with it measures 180° - 77° = 103° by the linear pair postulate.
Question 18 of 20 · Short Answer
Find the point on the x-axis that is equidistant from C(0, 4) and D(6, 2). Then explain why this point must lie on the perpendicular bisector of CD.
Let the point be (x, 0). Set the squared distances equal: x² + 16 = (x - 6)² + 4, so 16 = -12x + 40 and x = 2. The point (2, 0) is √20 from both C and D. By the converse of the perpendicular bisector theorem, a point equidistant from the endpoints lies on the perpendicular bisector. (Check: the midpoint is (3, 3), and (2, 0) - (3, 3) = (-1, -3) is perpendicular to CD, since (-1)(6) + (-3)(-2) = 0.)
Question 19 of 20 · Short Answer
The standard says the points on a perpendicular bisector of a segment are "exactly those" equidistant from the endpoints. Write the two if-then statements this phrase contains, and explain why proving only one of them is not enough.
Theorem: if a point is on the perpendicular bisector of AB, then it is equidistant from A and B. Converse: if a point is equidistant from A and B, then it is on the perpendicular bisector of AB. The first shows every point of the bisector qualifies. Only the second rules out equidistant points elsewhere. Together they show the bisector is the whole set of equidistant points, nothing more and nothing less.
Question 20 of 20 · Short Answer
Use the angle numbering of Diagram 1 with m ∥ n. m∠1 = (3x + 8)° and m∠8 = (5x - 36)°. Find x and m∠1, and justify each step with a theorem.
∠1 ≅ ∠5 by the corresponding angles theorem, and ∠5 ≅ ∠8 by the vertical angles theorem, so ∠1 ≅ ∠8 by the transitive property. Then 3x + 8 = 5x - 36, so x = 22 and m∠1 = 74°.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.CO.C.9 mean?
HSG.CO.C.9 means students must prove, not just use, theorems about lines and angles. The standard names vertical angles, the angles formed when a transversal crosses parallel lines, and the perpendicular bisector of a segment. A proof here is a chain of statements, each backed by a definition, a postulate or an earlier theorem.
Is HSG.CO.C.9 taught in Geometry or Algebra 1?
It is usually taught in high school Geometry, often in the first unit on proof. Students already used these angle facts informally in grades 7 and 8 (7.G.B.5 and 8.G.A.5). The new part is writing formal proofs.
What is the difference between a postulate and a theorem?
A postulate is accepted without proof, and a theorem is proved from postulates, definitions and earlier theorems. In this lesson the linear pair postulate and the parallel postulate are accepted. The vertical angles theorem is then proved from the linear pair postulate. Textbooks differ on which facts they treat as postulates, so check yours.
Why do we need to prove that vertical angles are congruent if it is obvious?
Because a picture shows one case and a proof covers every case. Two lines drawn nearly perpendicular make the fact look obvious, but a proof shows it holds for any angle, which later proofs rely on. It is also a short first proof that shows how reasons connect.
How do you prove that alternate interior angles are congruent?
Chain two earlier theorems. An alternate interior angle is vertical to an angle that corresponds to the other alternate interior angle. Vertical angles are congruent, corresponding angles are congruent when the lines are parallel, and the transitive property finishes the proof. A half-turn about the midpoint of the transversal segment is another way to prove it.
What is a common mistake in proofs about parallel lines?
Using the parallel-line theorems when the lines are not known to be parallel. The corresponding and alternate interior angles theorems start with "if the lines are parallel". A second common mistake is using the theorem being proved as one of its own reasons.
What does "exactly those" mean in the perpendicular bisector theorem?
It means the statement goes both ways. Every point on the perpendicular bisector is equidistant from the endpoints, and every point equidistant from the endpoints is on the perpendicular bisector. Students prove the theorem with SAS (or a reflection) and the converse with SSS.
Where is the perpendicular bisector theorem used later?
It explains why the compass construction of a perpendicular bisector works (HSG.CO.D.12). It also shows that the perpendicular bisectors of a triangle meet at a point equidistant from all three vertices, the circumcenter (HSG.CO.C.10 and HSG.C.A.3).
How are HSG.CO.C.9 proofs usually assessed?
Usually with a mix of angle calculations and proofs. Students find unknown angles with algebra and name the theorem used, complete or order the steps of a two-column proof, and write short paragraph proofs. On the digital SAT, angle relationships with parallel lines appear in the Geometry and Trigonometry domain, but formal proofs do not.
How can parents help with this standard at home?
Ask "How do you know?" after every answer. When your student says an angle is 65°, ask which theorem gives that. Road crossings, railroad tracks and window frames give real examples of transversals and parallel lines to talk about.
07
Related Standards
5 standards
These standards connect to HSG.CO.C.9: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.G.B.5Prerequisite
Use supplementary, complementary, vertical and adjacent angles to find unknown angles
