In plain English: HSG.CO.C.10 is the Common Core geometry standard that asks students to prove theorems about triangles. The official list includes the 180° angle sum, the congruent base angles of an isosceles triangle, the midsegment theorem and the fact that the three medians meet at one point. It is usually taught in high school Geometry.
Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.
Common Core State Standards for Mathematics · Domain: Congruence (CO) · Cluster: Prove geometric theorems Also written as HSG-CO.C.10 or G-CO.10 · Official standard
Students prove the four triangle theorems named in the standard: the interior angles of a triangle add to 180°, the base angles of an isosceles triangle are congruent, the segment joining the midpoints of two sides is parallel to the third side and half as long, and the three medians meet at one point. Each proof uses tools students already have: the parallel postulate and the angle pairs formed by a transversal, the SAS and SSS criteria, and coordinates with the midpoint and slope formulas.
The lesson separates checking from proving. Tearing the corners off one paper triangle, or computing one example, shows that a theorem is plausible. A proof has to work for every triangle, so students learn to name each reason and, in coordinate proofs, to use letters for the vertices instead of numbers. Numerical problems follow each proof so students can use the theorem as well as justify it.
Learning Objectives
By the end of this lesson, students will be able to:
Prove that the interior angles of any triangle add to 180°, using a line through one vertex parallel to the opposite side
Prove that the base angles of an isosceles triangle are congruent, using an auxiliary segment and a congruence criterion
Prove with general coordinates that a midsegment is parallel to the third side and half its length
Prove that the three medians of a triangle meet at one point, and locate that point from the vertices
Use these theorems to find unknown angles, lengths and coordinates
Prior Knowledge Required
Students should already be comfortable with:
Informal arguments about the angle sum of a triangle and angles formed by parallel lines and a transversal 8.G.A.5
Proving that alternate interior angles are congruent when a transversal crosses parallel lines HSG.CO.C.9
The SAS and SSS criteria for triangle congruence HSG.CO.B.8
The midpoint formula and the fact that parallel lines have equal slopes
Solving linear equations in one variable HSA.REI.B.3
"Maya measured the angles of five different triangles, and every time they added to 180°. She says that proves every triangle has an angle sum of 180°. Do you agree? What would convince you?"
Take a few answers. Draw out two points: measurements carry error (one student might get 178° or 182°), and five examples cannot cover every triangle. A proof is an argument that works for any triangle at once. Tell students they will prove four theorems today and use each one right away.
Direct Instruction25 minutes
Angle sum (Diagram 1). Given △ABC, draw the line ℓ through B parallel to AC; by the parallel postulate there is exactly one. ∠1 and ∠A are alternate interior angles for the parallel lines ℓ and AC cut by transversal AB, so they are congruent. For the same reason, with transversal BC, ∠3 ≅ ∠C. Angles 1, 2 and 3 together form a straight angle along ℓ, so m∠1 + m∠2 + m∠3 = 180°. Substituting gives m∠A + m∠ABC + m∠C = 180°.
Isosceles base angles. Given AB = AC. Let the bisector of ∠A meet BC at D. In △ABD and △ACD, AB = AC (given), ∠BAD ≅ ∠CAD (definition of angle bisector) and AD = AD (the same segment). By SAS, △ABD ≅ △ACD, so the corresponding angles ∠B and ∠C are congruent.
Midsegment theorem (coordinate proof). Place the triangle with B(0, 0), C(2c, 0) and A(2a, 2b), b ≠ 0. Doubling the coordinates avoids fractions. The midpoint of AB is M(a, b) and the midpoint of AC is N(a + c, b). MN and BC both have slope 0, and MN lies on the line y = b while BC lies on y = 0, so MN ∥ BC. MN = |(a + c) - a| = |c| and BC = |2c|, so MN = ½BC. Any triangle can be moved by a rigid motion into this position, so the proof covers every triangle.
Medians meet at a point (coordinate proof). Let the vertices be A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) and let G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3). The midpoint of BC is M = ((x₂ + x₃)/2, (y₂ + y₃)/2). The point two-thirds of the way from A to M is A + ⅔(M - A), whose x-coordinate is x₁ + ⅔((x₂ + x₃)/2 - x₁) = (x₁ + x₂ + x₃)/3, and the same for y. So G is on median AM. The formula for G treats the three vertices the same way, so the same computation puts G on the other two medians. The three medians meet at G, the centroid.
After each proof, work the matching example below. Point out that the examples use the theorems, while the proofs justify them. The fifth example also shows the 2:1 split that the coordinate proof produced: G is two-thirds of the way from each vertex to the opposite midpoint.
Triangle angle sum
The angles of a triangle measure (2x + 5)°, (3x - 10)° and (x + 35)°. Find each angle.
Equation: (2x + 5) + (3x - 10) + (x + 35) = 180, so 6x + 30 = 180 and x = 25. The angles are 55°, 65° and 60°
Isosceles base angles
In △PQR, PQ = PR and m∠P = 48°. Find m∠Q and m∠R.
Equation: The base angles are congruent, so m∠Q = m∠R = (180° - 48°) ÷ 2 = 66°
Midsegment length
D and E are the midpoints of AB and AC. DE = 2x - 1 and BC = 3x + 4. Find x, DE and BC.
Equation: BC = 2 · DE, so 3x + 4 = 2(2x - 1) = 4x - 2 and x = 6. DE = 11 and BC = 22
Midsegment on the coordinate plane (Diagram 2)
△ABC has A(2, 8), B(-4, 0), C(8, 2). M and N are the midpoints of AB and AC. Verify the midsegment theorem.
Equation: M(-1, 4) and N(5, 5). Slope MN = 1/6 and slope BC = 2/12 = 1/6, so MN ∥ BC. MN = √37 and BC = √148 = 2√37, so MN = ½BC
Medians meet at one point (Diagram 2)
△ABC has A(0, 0), B(10, 2), C(2, 10). Show that all three medians pass through G(4, 4).
Equation: Midpoints: (6, 6), (1, 5), (5, 1). Two-thirds of the way from each vertex to the opposite midpoint: (0, 0) + ⅔(6, 6) = (4, 4); (10, 2) + ⅔(-9, 3) = (4, 4); (2, 10) + ⅔(3, -9) = (4, 4)
Guided Practice15 minutes
Pairs complete a second proof of the isosceles base angles theorem, this time using the midpoint of the base instead of the angle bisector. Given △XYZ with XY = XZ, and M the midpoint of YZ.
Proof frame: base angles of an isosceles triangle using the midpoint of the base
Statement
Reason
1. XY = XZ
Given
2. M is the midpoint of YZ, so YM = ZM
____ (Definition of midpoint)
3. XM = XM
____ (A segment is congruent to itself)
4. △XYM ≅ △XZM
____ (SSS, from steps 1-3)
5. ∠Y ≅ ∠Z
____ (Corresponding parts of congruent triangles)
Then ask pairs to extend the proof: from the same congruence, why is ∠XMY ≅ ∠XMZ, and why does that make XM perpendicular to YZ? (The two angles are congruent and form a linear pair, so each is 90°.) Compare the two proofs as a class: one uses SAS with a bisector, the other SSS with a midpoint, and both reach the same theorem.
Independent Practice15 minutes
Students work alone. (1) A triangle has angles (4x)°, (5x - 6)° and (3x + 6)°. Find each angle. (12x = 180, so x = 15: 60°, 69° and 51°.) (2) In △XYZ, XY = XZ and m∠Y = 38°. Find m∠X. (m∠Z = 38°, so m∠X = 104°.) (3) △PQR has P(0, 0), Q(8, 6), R(10, 0). Find the midpoints of QP and QR and verify the midsegment theorem for the segment joining them. ((4, 3) and (9, 3): the segment is horizontal like PR, and its length 5 is half of PR = 10.) (4) For the same triangle, find the point where the medians meet. ((6, 2).)
Closure5-10 minutes
Exit ticket: (1) Name the postulate and the theorem about parallel lines that the angle sum proof depends on. (2) An isosceles triangle has base angles of 29°. Find the vertex angle. (122°.) (3) D and E are the midpoints of two sides of a triangle whose third side is 17 cm. How long is DE, and what else do you know about it? (8.5 cm, and DE is parallel to the third side.)
Differentiation Strategies
For Struggling Students
Give each proof as a frame with the statements filled in and the reasons missing, then remove statements as students gain confidence
Before a general coordinate proof, have students run the same steps on one numerical triangle and circle the numbers that will become letters
Keep a reason bank on the wall: parallel postulate, alternate interior angles, straight angle, definition of midpoint, SAS, SSS, corresponding parts
For Advanced Students
Ask students to prove the midsegment theorem without coordinates: extend DE to F with EF = DE, then show BCFD is a parallelogram
Ask students to prove that the centroid divides each median in the ratio 2:1 using only the midsegment theorem and similar triangles
Ask why the angle sum proof fails on a sphere, where there are no parallel lines (on a globe, a triangle can have three right angles)
Assessment Guidance
What to Look For
In each proof, check that every statement has a reason and that the reasons are ones already established: definitions, postulates or earlier theorems. Watch for circular steps, such as using the angle sum to prove the angle sum, and for claims read off the diagram ("the segments look parallel"). In coordinate proofs, the vertices must be written with letters, placed with a stated rigid motion, and the conclusion must follow from the algebra. In the computational problems, check that students name the theorem they used.
02
Classroom Activities
3 Activities
1
From Torn Corners to Proof
15 minPairs
Students first check the angle sum by hand, then turn the check into a proof. Tearing the corners suggests the key step: the three angles fit along a straight line.
Procedure
Each student cuts out a large triangle of a different shape (one acute, one obtuse per pair) and labels the corners a, b and c inside the triangle
Tear off the three corners and place their points together along the edge of a ruler. Record what you see
Draw the triangle again on paper, with the line through the top vertex parallel to the bottom side. Mark which torn corner fits into each of the three angles along that line
Write the proof in full, with a reason for each step
Discussion Questions
Why is the torn-corner activity a check and not a proof?
Which step of the proof matches the moment when the corners lined up along the ruler?
Modification for Distance Learning
In dynamic geometry software, students construct a triangle and the parallel line through one vertex, measure the five angles, and drag the vertices to see that the alternate interior angles stay equal.
2
Fold and Prove: Isosceles Triangles
20 minPairs
Students build isosceles triangles with a compass, fold them, and write a proof that explains what the fold shows.
Procedure
Draw a base segment of any length. With the compass set to 8 cm, draw arcs from both endpoints and join the intersection to the endpoints. Explain why the triangle is isosceles
Cut it out and fold it so the two 8 cm sides lie on top of each other. Record which angles land on each other and where the fold crosses the base
Write a proof that the base angles are congruent using the fold line as the angle bisector, as in Direct Instruction. Then, with your partner, check that the second proof from Guided Practice also works for your triangle
Discussion Questions
The fold is a reflection. Which points does it keep fixed, and which does it swap?
Where does the fold cross the base, and at what angle? Can you prove it?
Challenge Variation
Prove the converse: if ∠B ≅ ∠C in △ABC, then AB = AC. (One route: draw the bisector AD of ∠A. Triangles ABD and ACD share side AD and have congruent angles at A and at B and C, so by the angle sum their angles at D are congruent too. ASA with the shared side AD gives △ABD ≅ △ACD.)
3
Coordinate Proof Lab: Midsegments and Medians
20 minGroups of 3-4
Each student tests both theorems on one numerical triangle, then the group writes the general coordinate proofs and explains why they cover all four cards at once.
The 4 Triangle Cards
Card 1: A(0, 0), B(6, 0), C(0, 6)
Card 2: A(-2, -1), B(4, 1), C(1, 6)
Card 3: A(3, 9), B(-3, 0), C(9, -3)
Card 4: A(1, 1), B(7, 3), C(4, 8)
Procedure
Each student takes one card (groups of 3 skip Card 4). Find the midpoints M of AB and N of AC, then compare the slopes and lengths of MN and BC
Find the midpoint of each side, draw the three medians on graph paper, and compute the point ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3). Check that it is two-thirds of the way along each median
As a group, redo both computations with letters: B(0, 0), C(2c, 0), A(2a, 2b) for the midsegment, and A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) for the medians
Discussion Questions
Why is checking four cards still not a proof, while the letter version is?
Why was it helpful to write the midsegment vertices as 2a, 2b and 2c?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Proof That the Angles of a Triangle Add to 180°
Line ℓ is drawn through B parallel to AC. The alternate interior angles are congruent (∠1 ≅ ∠a and ∠3 ≅ ∠c), and angles 1, 2 and 3 form a straight angle, so the three angles of the triangle add to 180°.
Diagram 2: Midsegment and Medians on the Coordinate Plane
Left: in △ABC with A(2, 8), B(-4, 0), C(8, 2), the midsegment MN has the same slope as BC and half its length. Right: in △ABC with A(0, 0), B(10, 2), C(2, 10), the three medians meet at the centroid G(4, 4). Both are drawn to scale.
04
Homework Assignment
~30 min
HSG.CO.C.10 Homework: Proving Triangle Theorems
Directions: Give a reason for every statement in a proof. For numerical problems, name the theorem you use and show your equation.
Part 1: Angle Sum and Isosceles Triangles (Problems 1-2)
(a) The angles of a triangle measure (4x - 2)°, (x + 17)° and (2x + 4)°. Find x and each angle, and classify the triangle by its angles. (b) Prove that the interior angles of any triangle add to 180°, this time drawing the line through vertex A parallel to BC instead of the line through B. Name the postulate and the theorem about parallel lines you use.
(a) In △XYZ, XY = XZ, m∠Y = (5x - 8)° and m∠Z = (3x + 14)°. Find x, m∠Y and m∠X. (b) Prove that the base angles of an isosceles triangle are congruent using a third auxiliary segment: the altitude XH from X perpendicular to YZ. (Hint: use the Pythagorean Theorem in the two right triangles to show YH = ZH, then use a congruence criterion.)
Part 2: Midsegments (Problems 3-4)
△JKL has J(0, 6), K(-6, -2), L(4, -4). Let P be the midpoint of JK and Q the midpoint of JL. (a) Find P and Q. (b) Show that PQ ∥ KL using slopes. (c) Show that PQ = ½KL.
(a) D and E are the midpoints of AB and AC in △ABC. DE = x + 7 and BC = 4x - 10. Find x, DE and BC. (b) A triangle has sides 14 cm, 18 cm and 22 cm. Its three midsegments form a smaller triangle. Find the perimeter of the smaller triangle and explain which theorem you used.
Part 3: Medians and a New Proof (Problems 5-6)
△ABC has A(-3, 1), B(5, 7), C(7, -5). (a) Find the midpoint of each side. (b) Find G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3) and show that G lies on all three medians. (c) For the median from A, find AG and GM, where M is the midpoint of BC.
Prove that the three midsegments of any triangle divide it into four congruent triangles. (Hint: use the midsegment theorem to find the length of each midsegment, then use a congruence criterion.)
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Proof Logic
Every statement follows from earlier ones; no circular steps
One gap or unjustified step
Several gaps, or relies on the diagram
Reasons
Each reason is a definition, postulate or proved theorem, named correctly
Most reasons correct
Reasons missing or wrong
Computation
Equations, coordinates and lengths all correct
One error
Several errors or no work
Theorem Use
Names the theorem used in each numerical problem
Names some
Does not name theorems
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Select an answer for each multiple-choice question; the page tells you right away whether it is correct and why. Write out the short-answer proofs and computations before you reveal the model answer. Reset quiz starts over.
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 angles of a triangle measure 38° and 77°. What is the third angle?
Answer: A
180° - 38° - 77° = 65°. Choice B is the sum of the two given angles, not the third angle. Choice D subtracts from 360°, the angle sum of a quadrilateral.
Question 2 of 20 · Multiple Choice
In the proof of the triangle angle sum, a line through B is drawn parallel to AC. Which fact shows that the angle at A equals one of the angles at B?
Answer: C
AB is a transversal of the parallel lines, and the angle at A and the angle at B on the other side of AB are alternate interior angles. Choice B would assume the triangle is isosceles. Choice A does not apply, because the angles are not formed by two crossing lines at one point.
Question 3 of 20 · Multiple Choice
The vertex angle of an isosceles triangle measures 110°. What does each base angle measure?
Answer: B
The base angles are congruent and add to 180° - 110° = 70°, so each is 35°. Choice A forgets to split the 70° between the two base angles. Choice C divides 110° by 2.
Question 4 of 20 · Multiple Choice
Each base angle of an isosceles triangle measures 52°. What is the vertex angle?
Answer: D
180° - 2(52°) = 76°. Choice B subtracts only one base angle from 180°. Choice C comes from (180° - 52°) ÷ 2, which treats 52° as the vertex angle.
Question 5 of 20 · Multiple Choice
D and E are the midpoints of sides AB and AC of △ABC, and BC = 26 cm. What is DE?
Answer: A
By the midsegment theorem, DE = ½BC = 13 cm. Choice C doubles BC instead of halving it. Choice D halves twice.
Question 6 of 20 · Multiple Choice
D and E are the midpoints of AB and AC. DE = 3x - 2 and BC = 5x + 2. What is DE?
Answer: C
5x + 2 = 2(3x - 2) = 6x - 4, so x = 6, DE = 16 and BC = 32. Choice A is the value of x. Choice B is BC. Choice D drops the -2 when substituting x = 6 into 3x - 2.
Question 7 of 20 · Multiple Choice
△ABC has A(1, 2), B(7, 4), C(4, 9). Where do its medians meet?
Answer: B
The medians meet at ((1 + 7 + 4)/3, (2 + 4 + 9)/3) = (4, 5). Choice C divides the sums by 2 instead of 3. Choice A is the midpoint of BC, which is only one end of a median. Choice D divides the x-sum by 3 but the y-sum by 2.
Question 8 of 20 · Multiple Choice
△ABC has A(0, 4), B(-2, 0), C(6, 0). M and N are the midpoints of AB and AC. Which statement is true?
Answer: A
M(-1, 2) and N(3, 2), so MN is horizontal like BC, and MN = 4, which is half of BC = 8. Choice C gives the length of BC. Choice D is false because MN shares point M with AB, and parallel lines have no common point.
Question 9 of 20 · Multiple Choice
To prove the base angles theorem, a student draws the bisector AD of the vertex angle of △ABC, where AB = AC. Which criterion shows △ABD ≅ △ACD?
Answer: D
AB = AC (given), ∠BAD ≅ ∠CAD (bisector) and AD = AD (shared side): two sides and the included angle. Choice A would need BD = CD, which is not known yet. Choice C is not a congruence criterion.
Question 10 of 20 · Multiple Choice
The angles of a triangle are in the ratio 2 : 3 : 5. What is the largest angle?
Answer: C
Write the angles as 2k, 3k and 5k. Then 10k = 180, so k = 18 and the angles are 36°, 54° and 90°. Choice B uses 5k with k = 20, from 9k = 180, which leaves out one part of the ratio.
Question 11 of 20 · Multiple Choice
An exterior angle of a triangle is formed at vertex C. The two remote interior angles measure 45° and 70°. What is the exterior angle? (This follows from the angle sum theorem.)
Answer: B
The interior angle at C is 180° - 45° - 70° = 65°, and the exterior angle forms a linear pair with it: 180° - 65° = 115°. So the exterior angle equals 45° + 70°. Choice A gives the interior angle at C instead.
Question 12 of 20 · Multiple Choice
Why can a triangle not have two right angles?
Answer: D
By the angle sum theorem, the third angle would measure 180° - 90° - 90° = 0°, and a triangle cannot have a 0° angle. Choice B is false: acute triangles have no obtuse angle. Choice C is false: a 30-60-90 triangle is a right triangle that is not isosceles.
Question 13 of 20 · Multiple Choice
A triangle has sides 10 cm, 16 cm and 20 cm. What is the perimeter of the triangle formed by its three midsegments?
Answer: C
Each midsegment is half of the side it is parallel to: 5 + 8 + 10 = 23 cm, half the original perimeter of 46 cm. Choice A is the original perimeter. Choice B doubles it.
Question 14 of 20 · Multiple Choice
G is the point where the medians of △ABC meet, and AM is the median from A. AM = 18 cm. How long is AG?
Answer: B
The coordinate proof shows G is two-thirds of the way from each vertex to the opposite midpoint, so AG = ⅔(18) = 12 cm and GM = 6 cm. Choice A assumes G is the midpoint of the median. Choice C gives GM instead of AG.
Question 15 of 20 · Short Answer
Prove that the interior angles of any triangle add to 180°. Give a reason for each step.
Draw the line through one vertex parallel to the opposite side and use alternate interior angles. In △ABC, draw line ℓ through B parallel to AC (parallel postulate). With transversal AB, the angle between ℓ and BA is congruent to ∠A (alternate interior angles). With transversal BC, the angle between BC and ℓ is congruent to ∠C. Those two angles and ∠ABC form a straight angle along ℓ, so their measures add to 180°. Substituting, m∠A + m∠ABC + m∠C = 180°.
Question 16 of 20 · Short Answer
Prove that the base angles of an isosceles triangle are congruent.
Split the triangle into two congruent triangles. In △ABC with AB = AC, draw the bisector of ∠A meeting BC at D. Then AB = AC (given), ∠BAD ≅ ∠CAD (definition of bisector) and AD = AD (same segment), so △ABD ≅ △ACD by SAS. Corresponding parts of congruent triangles are congruent, so ∠B ≅ ∠C. (Using the midpoint of BC and SSS also works.)
Question 17 of 20 · Short Answer
△ABC has A(4, 6), B(0, 0), C(10, 0). Verify the midsegment theorem for the segment joining the midpoints of AB and AC.
The midpoints are (2, 3) and (7, 3); the segment is parallel to BC and half as long. Both points have y = 3, so the segment is horizontal, like BC on the x-axis, and the two lines are parallel (y = 3 and y = 0). Its length is 7 - 2 = 5, and BC = 10, so the midsegment is half of BC.
Question 18 of 20 · Short Answer
△ABC has A(0, 0), B(12, 0), C(6, 9). Show that all three medians pass through one point, and name the point.
The medians meet at (6, 3). The midpoints are (9, 4.5) on BC, (3, 4.5) on AC and (6, 0) on AB. The median from C to (6, 0) is the vertical line x = 6, which contains (6, 3). The median from A to (9, 4.5) lies on y = x/2, and 3 = 6/2. The median from B to (3, 4.5) lies on y = -(x - 12)/2, and -(6 - 12)/2 = 3. So (6, 3), which is ((0 + 12 + 6)/3, (0 + 0 + 9)/3), is on all three.
Question 19 of 20 · Short Answer
In △PQR, PQ = PR, m∠Q = (4x + 6)° and m∠R = (6x - 14)°. Find x, m∠Q and m∠P.
x = 10, m∠Q = 46° and m∠P = 88°. The base angles of an isosceles triangle are congruent, so 4x + 6 = 6x - 14, 20 = 2x and x = 10. Then m∠Q = m∠R = 46°, and by the angle sum, m∠P = 180° - 92° = 88°.
Question 20 of 20 · Short Answer
A carpenter builds an A-frame shelf. The two legs meet at the top, and their feet stand 3.6 m apart on the floor. A crossbar joins the midpoints of the two legs. How long is the crossbar, and how does the carpenter know it is level with the floor?
The crossbar is 1.8 m long, and it is parallel to the floor. The legs and the floor form a triangle, and the crossbar joins the midpoints of two of its sides, so it is a midsegment. By the midsegment theorem it is parallel to the third side (the floor) and half its length: ½(3.6) = 1.8 m.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.CO.C.10 mean?
HSG.CO.C.10 means students prove theorems about triangles, not just use them. The standard lists four examples: the angle sum is 180°, isosceles base angles are congruent, a midsegment is parallel to the third side and half as long, and the medians meet at one point. Teachers may add others, such as the exterior angle theorem.
Which theorems does HSG.CO.C.10 require?
The standard says "theorems include" the four listed above, so those four are the core. Courses often add related results that follow from them, like the exterior angle theorem, the converse of the base angles theorem, and the fact that the centroid divides each median in a 2:1 ratio.
Is HSG.CO.C.10 taught in Geometry or Algebra 1?
It is a high school Geometry standard. It usually comes after students have proved theorems about lines and angles (HSG.CO.C.9) and learned the congruence criteria (HSG.CO.B.8), because the triangle proofs depend on both.
Do the proofs have to be two-column proofs?
No. The Common Core does not require a format. Two-column proofs, paragraph proofs, flow charts and coordinate proofs are all valid, as long as each step has a reason and the argument covers every triangle, not a single example.
Why is measuring angles not a proof that they add to 180°?
Because measurement is approximate and only covers the triangles you measured. A proof uses definitions, postulates and earlier theorems to show the result holds for every triangle. Measuring or tearing corners is useful for making a conjecture, and the proof comes after.
What is a midsegment of a triangle?
A midsegment is the segment joining the midpoints of two sides. Every triangle has three. The midsegment theorem says each one is parallel to the third side and half as long, so the three midsegments form a triangle with half the perimeter of the original.
Can coordinate proofs be used for HSG.CO.C.10?
Yes, as long as the coordinates are general. Placing a triangle at B(0, 0), C(2c, 0) and A(2a, 2b) represents any triangle, since a rigid motion can move any triangle into that position. A proof with specific numbers only checks one case. Coordinate methods are also the focus of HSG.GPE.B.4.
What mistakes do students make in triangle proofs?
A common error is using the result being proved as a reason, for example using the angle sum inside its own proof. Others are assuming a fact because the diagram looks that way, skipping the reason for an auxiliary line (such as why the parallel line exists), and writing a coordinate proof with numbers instead of variables.
Does the angle sum theorem work on a sphere?
No. The proof depends on the parallel postulate, which does not hold on a sphere. On a globe, a triangle with one vertex at the North Pole and two on the equator can have three right angles, for a sum of 270°. This is a good discussion question for advanced students.
How does HSG.CO.C.10 connect to later topics?
The angle sum supports every later angle problem, and the midsegment theorem previews the triangle proportionality theorem in the similarity unit (HSG.SRT.B.4). The centroid appears again in coordinate geometry and, outside math class, as the balance point of a flat triangular plate.
07
Related Standards
5 standards
These standards connect to HSG.CO.C.10: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.A.5Prerequisite
Informal arguments about triangle angle sums and parallel lines cut by a transversal
