**What is a Median Triangle?**

For any **Triangle ABC, **a median
triangle is a triangle whose sides have the lengths of the three
medians of **Triangle ABC**. A construction of such a triangle
is demonstrated below.

**A Construction**

Start with **Triangle ABC** and its
medians, **Segments AD**, **BE**, and **CF **and centroid
**G.**

.

**Segment FC**
will be the first side of the median triangle. To create the second
side, **Segment BE** will be translated such that the image
of** E** will coincide with** C**. The resulting segment
is **Segment HC**.

Using a similar process, **Segment AD**
will be translated such that the image of **A** will coincide
with** F**.

It appears that the image of **Segment
AD** is **Segment FH**. There is no proof, however, that
the image of **D** is indeed **H** from the first translation.
Until this is proved, the image of **D** will be called **H'**.
If **H** and **H**' are the same point, then the images
of** B** and** D** coincide, resulting in a triangle. By
examining vectors, this can be established.

**A Proof Verifying the Translation
of the Medians Results in a Triangle**

Three vectors have been established in the above figure:

**Vector a**,
representing the direction from **C **to **D**, whose magnitude
is half the length of **Segment BC**.

**Vector b**,
representing the direction from **A** to **F**, whose magnitude
is half the length of **Segment AB**. Note that **Vector b**
could also extend from **E** to **D**, as **Segment DE**
is the midsegment of **Triangle ABC**.

**Vector c**,
representing the direction from **E** to **C**, whose magnitude
is half the length of **Segment AC**.

Note that: **Vector a + Vector b = Vector
c**

This leads to **Vector b = -Vector a
+ Vector c**

In the figure below, the translation
of **Segment BE** to **Segment HC** is represented by the
**Vector c**. Similarly, the translation of **Segment AD**
to **Segement BH'** is represented by **Vector b**.

To arrive at **H** from **D**,
one could travel along **Vector b**.

To arrive at **H**' from **D**,
one could travel along the opposite of **Vector a** to **B**
and then along **Vector C** to **H**'.

From the second vector equation above,
the path from **D** to **H** and the path from **D**
to **H**' arrive at the same place. Thus, **H** and **H**'
must be the same point and the construction creates the median
triangle, **Triangle CFH**. QED : )

**The Area of the Median Triangle**

Click Here. This link will open a Geometer's Sketchpad file containing the median triangle construction. You can use this to explore the relationships between a triangle and its median triangle. One such conjecture is below.

Conjecture: The area of a median triangle is 3/4 the area of the triangle from which it was constructed.

**Given: Triangle ABC and its median
triangle, Triangle CFH.**

**Prove: The area of Triangle CFH is
three fourths of the area of Triangle ABC.**

For a proof of this conjecture, refer
to the figure below. Let the area of the original (orange) triangle
be **A**. To verify the above conjecture, the median triangle
(green) must have an area of **3A/4**.

There's a lot going on here. Some things can be established right away.

Since the medians of a triangle divide
it into six smaller triangles of equal area, the areas of **Triangles
AGE**, **EGC**, **CGD**, **DGB**, **BGF**, and **FGA**
are all **A/6**.

Since **Segment FD** is the midsegment
of **Triangle ABC**, the area of **Triangle BFD** is one
fourth of the area of **Triangle ABC**, **A/4**.

Since **F** is the midpoint of **Segment
AB** and **Line FH** is parallel to **Segment AD** (by
construction), **Segement FK** is the midsegment of **Triangle
BAD**. So, **Segments DK** and **KB** are congruent making
the areas of **Triangles BKF** and **DKF** equivalent and,
since the triangles partition **Triangle BFD**, equal to **A/8**.

The **Triangles BGF** and **DGB**
form **Qudrilateral BFGD**, whose area is **A/3**. If **Triangle
BFD**, with area **A/4**, is removed, **Triangle FGD**
remains with area **A/12**.

The diagram below shows the area of the
portion (colored dark green) of the median triangle (**CFH**)
that overlaps the original triangle (**ABC**). The sum of the
areas in the dark green regions is **3A/8**. Note that, in
order for the median triangle to have an area of **3A/4**,
the light green, non-overlapping portion must have the same area
as the dark green portion. This can be established by showing
that **Segments FK** and **KH** are congruent.

One way to show that **Segments FK**
and **KH** are congruent is to show that **Quadrilateral FBHD**
is a parallelogram. If this is the case, the diagonals of the
parallelogram will bisect each other, yielding the desired congruent
segments.

Since **Segments DH** and **FB**
are parallel to each other (by construction), **Angle FHD**
is congruent to **Angle BFH** as they are alternate interior
angles. Similarly, **Angle HDB** is congruent to **Angle FBD**.
Due to these congruent angle pairs and congruent segements **BK**
and **KD** (from above), **Triangles BKF** and **DKH**
are congruent. As a result, corresponding **Segments BF** and
**DH** must be congruent.

Since segments **BF** and **DH**
are both parallel and congruent, **Quadrilateral FBHD** is
a parallelogram, making **Segment FK** congruent to **Segment
KH**. Thus, **Triangle HKC** has the same area of **Triangle
FKC**, **3A/8**. Together, they have a combined area of **3A/4**,
three quarters of the area of **Triangle ABC**. QED : )

What other conjectures can be made about median triangles? Can they be proved?