Copyright

Proving the Triangle Midsegments Theorem

Instructor: Laura Pennington

Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.

The Triangle Midsegment Theorem is extremely useful in real-world applications. This lesson will state the Triangle Midsegment Theorem, apply it to an example, and then provide a proof of the theorem.

Triangle Midsegment Theorem

Suppose there are four roads in your neighborhood, such that three of the roads form a triangle, and the fourth road (Smith Street) connects two of the other roads at their midpoints.


trimidthm1


Did you know there is an actual theorem about Smith Street? The Triangle Midsegment Theorem states that the line segment connecting the midpoints of any two sides of a triangle will satisfy the following properties:

  1. The line segment will be parallel to the third side.
  2. The length of the line segment will be one-half the length of the third side.

In terms of the roads, this tells us that Smith Street is parallel to Bradford Road, and that Smith Street is one-half the distance of Bradford Road. Neat! We can use the theorem to determine that Smith Street has length 2 km, since 1/2 × 4 = 2.

The proof of this theorem involves proving both of the properties using a series of logical deductions along with several other theorems and properties. We definitely want to see this proof, so let's first familiarize ourselves with some facts and properties that we'll need to use to carry out the proof.

Facts for the Proof

Get ready for a lot of information at once. However, it will be not only useful, but essential to have these to look at when we do the proof. If you're unfamiliar with why these facts are true, don't worry. They're just here for reference. We're more interested in the facts themselves.


trimidthm2


In a triangle ABC, if we connect the midpoints, D and E, of any two sides, say AB and BC, as shown in the image, then the following facts are true:

  1. AD = BD and CE = BE, because D and E are the midpoints of AB and BC, respectively.
  2. Side Angle Side Triangle Similarity Theorem: Triangles are similar if the ratio between two corresponding sides is equal to the ratio between another two corresponding sides, and the angles between those corresponding sides have the same measure.
  3. Similar triangles have corresponding angles of equal measure, and corresponding sides are proportional, or create the same ratio.
  4. Converse of the Corresponding Angles Postulate: If we draw a line, called a transversal, through any two lines, and it creates two corresponding angles that have equal measure, then the two lines are parallel.

Okay, I told you it was a lot of information all at once, but we're going to be very glad that we have it handy when we do this proof. Speaking of which, we're all set! Let's do this!

Proof of the Triangle Midsegment Theorem

Given a triangle ABC, let's draw a line segment connecting the midpoints of two of the sides, say AB and BC.


trimidthm3


To prove the Triangle Midsegment Theorem, we need to show two things:

  1. DE is one-half of AC. That is, DE = (1/2)AC.
  2. DE is parallel to AC.

We'll start with part one. Because D is the midpoint of AB, we have that AD = DB. Therefore,

  • AB = AD + DB = DB + DB = 2DB

Similarly, since E is the midpoint of BC, we have that BE = EC, and

  • BC = BE + EC = BE + BE = 2BE

Therefore, we have the following:

  • DB / AB = BE / BC = 1/2

Consider the ΔABC and ΔDBE. We just showed that the ratio between corresponding sides DB and AB and between corresponding sides BE and BC are both equal to 1/2. Also, notice that these two triangles share the angle B, so we know the measure of the angle between these two sets of corresponding sides have equal measure. Therefore, by the Side Angle Side Theorem for Similar Triangles, we have that ΔABC is similar to ΔDBE.

Since ΔABC is similar to ΔDBE, we have that all of the corresponding sides of these two triangles are proportional and create the ratio 1/2.

To unlock this lesson you must be a Study.com Member.
Create your account

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use Study.com

Become a Study.com member and start learning now.
Become a Member  Back
What teachers are saying about Study.com
Try it risk-free for 30 days

Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it risk-free for 30 days!
Create an account
Support