# Special Lines in Triangles - Medians

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

Worksheets for Geometry, Module 1, Lesson 30

Student Outcomes

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

Special Lines in Triangles - Medians

Classwork

Opening Exercise

In β³ π΄π΅πΆ to the right, π· is the midpoint of π΄π΅, πΈ is the midpoint of π΅πΆ, and πΉ is the midpoint of π΄πΆ. Complete each statement below.

π·πΈ is parallel to ____ and measures ____ the length of ____. π·πΉ is parallel to ____ and measures ____ the length of ____. πΈπΉ is parallel to ____ and measures ____ the length of ____.

Discussion

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.

Example 1

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

Given: β³ π΄π΅πΆ with π·, πΈ, and πΉ the midpoints of sides Μπ΄π΅ΜΜΜ, π΅πΆΜΜΜΜ, and π΄πΆΜΜΜΜ, respectively
Prove: The three medians of β³ π΄π΅πΆ meet at a single point.

(1) Draw midsegment π·πΈ. Draw π΄πΈ and π·πΆ; label their intersection as point πΊ.
(2) Construct and label the midpoint of π΄πΊ as point π» and the midpoint of πΊπΆ as point π½.
(3) π·πΈ β₯ π΄πΆ,
(4) π»π½ β₯ π΄πΆ,
(5) π·πΈ β₯ π»π½,
(6) π·πΈ = 1/2 π΄πΆ and π»π½ = 1/2π΄πΆ,
(7) π·πΈπ½π» is a parallelogram.
(8) π»πΊ = πΈπΊ and π½πΊ = π·πΊ,
(9) π΄π» = π»πΊ and πΆπ½ = π½πΊ,
(10) π΄π» = π»πΊ = πΊπΈ and πΆπ½ = π½πΊ = πΊπ·,
(11) π΄πΊ = 2πΊπΈ and πΆπΊ = 2πΊπ·,
(12) We can complete Steps (1)β(11) to include the median from π΅; the third median, π΅πΉ, passes through point πΊ, 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.

Example 2

In β³ π΄π΅πΆ, the medians are concurrent at πΉ. π·πΉ = 4, π΅πΉ = 16, π΄πΊ = 30. Find each of the following measures.
a. πΉπΆ =
b. π·πΆ =
c. π΄πΉ =
d. π΅πΈ =
e. πΉπΊ =
f. πΈπΉ =

Example 3

In the figure to the right, β³ π΄π΅πΆ is reflected over π΄π΅ to create β³ π΄π΅π·. Points π, πΈ, and πΉ are midpoints of π΄π΅, π΅π·, and π΅πΆ, respectively. If π΄π» = π΄πΊ, prove that ππ» = πΊπ.

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