A few facts about the medians of a triangle:

1. Each median divides the triangle in half.

2. The three medians are concurrent at a point called the centroid.

3. The centroid is two-thirds the distance from a vertex to the midpoint of the side opposite that vertex.

(for 2 and 3)GSP Verification

(for 2 and 3)PROOF

Here is a picture of a **triangle**, its medians, centroid
(G), and **triangle of
medians** using the lengths of the medians.
Notice that I used one of the medians (**segment Aa**) as one of the
sides of the triangle of medians. This causes the triangle of
medians to overlap with the original triangle.

How might we construct the **triangle of medians**? How do I make sure that **AP = cC** and **aP = Bb**?

The general idea is to use one of the medians
as one of the sides of your **triangle
of medians**.

By constructing a line through **point**
**a** that is parallel to **segment
Bb** and a line through **point b**
that is parallel to **side BC**, you can create a **parallelogram
BbPa** where **aP =
Bb.**

Similarly, by constructing a line through **point
A** that is parallel to **segment
cC **and** **a line through **point
C** that is parallel to **side AB, **you can construct**
parallelogram AcCP** where **AP
= cC.**

If you would like to see these lines used to
construct the **triangle
of medians**, double-click on the ** Construction
lines** button in the sketch.

The **triangle
of medians** does not appear to be congruent
or similar to the original triangle. However, it is worthwhile
to compare the areas and perimeters. According to the measurements,
it seems that there is ** no constant ratio of the perimeters**.
However, the ratio of the areas is a constant. In fact, the calculation
reveals that

Let's prove it:

First of all, notice how part of the **triangle of medians **overlaps a part of the **original triangle**. In fact,
it overlaps a certain fraction of the **original triangle**. Recall
that **median Aa **divides the triangle's area in half. Notice that **D AaJ** lies on most of **D AaC**, which is half the area
of the **original triangle**. How much of that half of the triangle is it???

Perhaps it would be easier to consider the
"leftover" part, **D CJa**. Recall that **segment** **aJ || segment Bb**. Therefore, **D CJa ~
D CbB. D CbB **is created
by a median; therefore, the area of **D CbB
**is half the area of the original triangle.
Since **point a** is the midpoint of **side BC**, then the
sides and height of **D CJa** are half the length of the sides and height of **D CbB**.

So,

Area of D CJa = (1/2)* [ (half of base of D CbB)*(half of height of D CbB) ]

= (1/2)* [ (1/4)*(base of D CbB)*(height of D CbB) ]

= (1/4)* [ (1/2)*(base of D CbB)*(height of D CbB) ]

= (1/4)*(area of D CbB)

= (1/4)*(area of D AaC) since D AaC is also created by a median and is therefore 1/2 the area of the original triangle.

Since (**area
of ****D
CJa) **+
(**area of** **D
AaJ)** = (**area of ****D AaC)**, then **D
AaJ
**must be** 3/4** the area of **D AaC**. Since the area of **D AaC** is already **1/2**
the area of the **original
triangle**, then **D AaJ** is **(3/4) of (1/2)** of the area of the **original triangle**.
That is, **D
AaJ**
is **3/8** the area of the **original triangle**.

Is **D AaJ** half the area of the **triangle of medians**? If
it is, then that would mean that the rest of the **triangle of medians**, **D
AJP**,
is also **3/8 **the area of the** ****original triangle**. Therefore,
the entire** triangle
of medians** would be (3/8 + 3/8) = 6/8
= 3/4 the area of the **original
triangle**.

Consider this picture:

**Segment bP || segment BC** (segment bP is part of the line parallel to side BC).
Therefore,

m< JaC = m< JPb(alternate interior angles)

m< CJa = m< bJP(vertical angles).

Also, **bP = Ba **(opposite sides of a parallelogram
are congruent) and **Ba = ac **(since a is the midpoint of
**BC**)

Therefore,

bP = ac.

**D JaC is congruent
to D JPb** (ASA
Congruence). So **Ja = JP.**

This means that **point J** is the midpoint
of **side aP** of the **triangle
of medians**. **Segment AJ** is a
median that divides the area of the **triangle of medians **in half. Therefore,

the area of the triangle of medians

= (the area of D AaJ) + (the area of D AJP)

= (3/8 the area of D ABC) + (3/8 the area of D ABC)

= 3/4 the area of D ABC.