the midpoints of a triangle common core geometry homework

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.

There isn't much to interpret here - we are to prove the listed relationships. As I have looked at these relationships I would say that not all of them need to be done at this point of the year. There are a variety of ways to prove these relationships - informally through symmetry or formally through transformations, congruent triangles or coordinates. I believe there are a few other concepts that are not specifically stated in this list but still need to be discussed such as the exterior angle theorem or the triangle inequality theorem.

DIRECTLY IMPLIED SKILLS

The student will be able to prove and apply that the sum of the interior angles of a triangle is 180°.

The student will be able to prove and apply that the base angles of an isosceles triangle are congruent.

The student will be able to prove and apply the midsegment (midline) of triangle theorem.

The student will be able to prove that the medians of a triangle meet at a point, a point of concurrency.

INDIRECTLY IMPLIED SKILLS

The student will be able to prove and apply that the exterior angle theorem.

The student will be able to determine the conditions for forming a triangle, when given three lengths.

THE BIG IDEA

All polygons can be divided into triangles – thus the proof and use of properties and relationships of triangles is essential to geometric study. 

The more we know about triangles -the more we know about all polygons. The listed relationships represent a few of the key theorems that get used later to build on other concepts.

TRAPS & PITFALLS

Proof.... is a trap!! Just kidding - that is what most students feel about them. I think a trap might be feeling that proofs must be presented in a two column organizational manner. I have been that teacher for 15+ years and now I find myself encouraging students to use more paragraph or flowchart proofs.

When our definition of congruent triangles is developed through transformations, then we need to develop proofs using that train of thought. The power of transformations is that they are visual and students connect with them easier than all of the technical jargon of the past. Expressing logical arguments in a written form, describing a process and a reason is a very powerful way to help students succeed with proof.

It takes time to teach students to language to use to match their visual understanding but the core unpinning logic is there because they see the transformations that would help move the proof along… this is a journey.. a new journey but I feel that more students are connected to proof through transformational point of view.

PAST CONNECTIONS

Students need to have a strong grasp on the definitions and properties of isometric transformations so they can establish relationships through transformations. They also need to know the newly established triangle congruence criteria so that they can be used in proof. They also need to know the relationships relating to parallel lines and a transversal. Notice that the early part of this unit has all been preparatory material so that students can use these concepts to prove new ideas and relationships.

FUTURE CONNECTIONS

These relationships that are specifically mentioned in this objective all become tools for later connections. The more we prove the more we can use to find other new relationships.

MY REFLECTIONS (over line l )

Proof has always been a difficult area to teach successfully. Before common core, textbooks always introduce conditional statements, properties of equality, algebraic proofs, etc.... all in preparation to prove triangles to be congruent. This was always a difficult journey for many students because of their limited exposure to logical reasoning.  Well the core curriculum changes that quite drastically. Relationships are not connected through a transformational approach which is visual and approachable.  The definitions have been altered to reflect congruence as a mapping between two shapes…. Yes I am still working out what do these things look like and how to write them correctly but I must admit I have way more students connected to the process and willing to try because they see it.

Again as I mentioned in a previous reflection, to aide in helping students create proofs I have developed a draft, revision and final approach.  I have them create proofs, have their peers read them and given them feedback and then attempt the writing again to clarify missed ideas or to fix the wording.  This has been very successful!!

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Triangle Midsegment Theorem

There are so many rules regarding triangles to learn about. Once we understand these rules, we can solve triangles and many other geometrical problems with a few easy steps. One of the most useful tricks is the triangle midsegment theorem. But wait a second.. What exactly is a midsegment? How does this theorem work? Let's find out:

The triangle midsegment theorem, explained

In the world of math, we can rely on theorems because other mathematicians have already proven that they are correct. Once proven, theorems become rules that we can apply to many different situations.

The triangle midsegment theorem states that:

• A midsegment connecting two sides of a triangle is parallel to the third side
• The midsegment is half as long as the third side

Remember that a midsegment is simply a line that connects the midpoints of two triangle sides. You might also remember that a midpoint is the halfway point of each triangle side. We must draw our midsegments between these two midpoints. This is what a midsegment looks like:

As we can see, points D and E are both midpoints, and they are both halfway between points A-B and A-C, respectively. We can call this midsegment DE.

Visualizing the triangle midsegment theorem

Let''s take a closer look at our midsegment:

We can see that both distances on either side of D and E are equal. If we know this is true, then we can apply the midsegment triangle theorem. We can also put this theorem into a handy formula:

If AD = DB and AE = EC , then DE ∥ BC and DE = 1 2 BC . Note that ∥ is the symbol for parallelism.

Using the triangle midsegment theorem to solve problems

Now let's put our knowledge of the triangle midsegment theorem to good use. Consider the following diagram:

We can see that P and Q are the midpoints of sides AB and AC, respectively. We know this because the values on each side of P and Q are equal. This lets us know that PQ is the midsegment with an unknown length (x).

We also know that line BC has a value of 6.

What is the length of the midsegment in this diagram? Finding the answer is simple if we apply the triangle midsegment theorem:

PQ = 1 2 BC

In other words, all we need to do is divide line BC by 2: 1 2 ( 6 ) = 3

The midsegment is exactly 3 units in length.

How do we know that the triangle midsegment theorem is correct?

Whenever we face a difficult geometrical exercise, it always helps to construct triangles within other shapes to find our answers.

We can prove the validity of the triangle midsegment theorem with the same strategy. If we construct a triangle with vertices of all three midpoints of a triangle, we are left with one triangle in the middle. We also get three triangles surrounding this central triangle.

Try it yourself. Even with a quick glance, you will see that the midsegment is equal to two halves of the third side.

Topics related to the Triangle Midsegment Theorem

30-60-90 Triangles

Triangle Angle Bisector Theorem

Flashcards covering the Triangle Midsegment Theorem

Advanced Geometry Flashcards

Common Core: High School - Geometry Flashcards

Practice tests covering the Triangle Midsegment Theorem

Common Core: High School - Geometry Diagnostic Tests

Advanced Geometry Diagnostic Tests

Help your student understand the triangle midsegment theorem

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Special Lines in Triangles - Medians

Related Topics: Lesson Plans and Worksheets for Geometry Lesson Plans and Worksheets for all Grades More Lessons for Geometry Common Core For Geometry

New York State Common Core Math Geometry, Module 1, Lesson 30

Worksheets for Geometry

Student Outcomes

• Students examine the relationships created by special lines in triangles, namely medians.

Opening Exercise

In △ 𝐴𝐵𝐶 to the right, 𝐷 is the midpoint of 𝐴𝐵, 𝐸 is the midpoint of 𝐵𝐶, and 𝐹 is the midpoint of 𝐴𝐶. Complete each statement below.

𝐷𝐸 is parallel to ____ and measures ____ the length of ____. 𝐷𝐹 is parallel to ____ and measures ____ the length of ____. 𝐸𝐹 is parallel to ____ and measures ____ the length of ____.

In the previous two lessons, we proved that (a) the midsegment of a triangle is parallel to the third side and half the length of the third side and (b) diagonals of a parallelogram bisect each other. We use both of these facts to prove the following assertion:

All medians of a triangle are ____. That is, the three medians of a triangle (the segments connecting each vertex to the midpoint of the opposite side) meet at a single point. This point of concurrency is called the ____, or the center of gravity, of the triangle. The proof also shows a length relationship for each median: The length from the vertex to the centroid is ____ the length from the centroid to the midpoint of the side.

Provide a valid reason for each step in the proof below.

Given: △ 𝐴𝐵𝐶 with 𝐷, 𝐸, and 𝐹 the midpoints of sides ̅𝐴𝐵̅̅̅, 𝐵𝐶̅̅̅̅, and 𝐴𝐶̅̅̅̅, respectively Prove: The three medians of △ 𝐴𝐵𝐶 meet at a single point.

(1) Draw midsegment 𝐷𝐸. Draw 𝐴𝐸 and 𝐷𝐶; label their intersection as point 𝐺. (2) Construct and label the midpoint of 𝐴𝐺 as point 𝐻 and the midpoint of 𝐺𝐶 as point 𝐽. (3) 𝐷𝐸 ∥ 𝐴𝐶, (4) 𝐻𝐽 ∥ 𝐴𝐶, (5) 𝐷𝐸 ∥ 𝐻𝐽, (6) 𝐷𝐸 = 1/2 𝐴𝐶 and 𝐻𝐽 = 1/2𝐴𝐶, (7) 𝐷𝐸𝐽𝐻 is a parallelogram. (8) 𝐻𝐺 = 𝐸𝐺 and 𝐽𝐺 = 𝐷𝐺, (9) 𝐴𝐻 = 𝐻𝐺 and 𝐶𝐽 = 𝐽𝐺, (10) 𝐴𝐻 = 𝐻𝐺 = 𝐺𝐸 and 𝐶𝐽 = 𝐽𝐺 = 𝐺𝐷, (11) 𝐴𝐺 = 2𝐺𝐸 and 𝐶𝐺 = 2𝐺𝐷, (12) We can complete Steps (1)–(11) to include the median from 𝐵; the third median, 𝐵𝐹, passes through point 𝐺, which divides it into two segments such that the longer part is twice the shorter. (13) The intersection point of the medians divides each median into two parts with lengths in a ratio of 2:1; therefore, all medians are concurrent at that point.

The three medians of a triangle are concurrent at the , or the center of gravity. This point of concurrency divides the length of each median in a ratio of ; the length from the vertex to the centroid is the length from the centroid to the midpoint of the side.

In △ 𝐴𝐵𝐶, the medians are concurrent at 𝐹. 𝐷𝐹 = 4, 𝐵𝐹 = 16, 𝐴𝐺 = 30. Find each of the following measures. a. 𝐹𝐶 = b. 𝐷𝐶 = c. 𝐴𝐹 = d. 𝐵𝐸 = e. 𝐹𝐺 = f. 𝐸𝐹 =

In the figure to the right, △ 𝐴𝐵𝐶 is reflected over 𝐴𝐵 to create △ 𝐴𝐵𝐷. Points 𝑃, 𝐸, and 𝐹 are midpoints of 𝐴𝐵, 𝐵𝐷, and 𝐵𝐶, respectively. If 𝐴𝐻 = 𝐴𝐺, prove that 𝑃𝐻 = 𝐺𝑃.

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Geometry, Common Core Style

Monday, december 14, 2015, lesson 11-5: the midpoint connector theorem (day 71).

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Geometry: Common Core (15th Edition)

By charles, randall i., chapter 5 - relationships within triangles - 5-1 midsegments of triangles - practice and problem-solving exercises - page 288: 9, work step by step, update this answer.

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