**The Medial Triangle**

For any given Triangle **ABC**, its medial
triangle is constructed by finding the midpoints of its three
sides (Segments **AB**, **BC**, and **AC**) and constructing segments to join these midpoints.
In the figure below, Triangle **DEF** is the medial triangle of Triangle **ABC**.

An important property of the medial triangle
is that it divides the original triangle into four smaller, congruent
triangles that are each similar to the original triangle. Thus
**Triangles AFD**, **DEB**, **FCE**, and **EDF** are
congrent to each other and similar to **Triangle ACB**.

**Can you prove this conjecture ?(You
may find the triangle midsegment theorem helpful)**

**What is the ratio of the corresponding
side lengths between triangle ACB and the smaller triangles?**

The examination of the triangle centers of any triangle and its medial triangle reveals some interesting relationships. For a quick review of the four triangle centers, Click Here. Below are some conjectures regarding the centers of medial triangles.

**Conjecture #1: The centroid
of a triangle is also the centroid of its medial triangle.**

To begin considering this hypothesis, examine the figure below.

It certainly appears that this is the case. Since appearances can often be deceiving in geometry, a proof of this conjecture is offered below.

**Given:**
**Triangle DEF** is the medial triangle of **Triangle ABC**;
**N** is the centroid of **Triangle ABC**.

**Prove:**
**N** is the centroid of **Triangle DEF**.

If **N** is the centroid of **Triangle
DEF**, then **Segements EL, DM, **and **FK** are medians
of **Triangle DEF**. If this is the case, then **K**,**
L**,** **and** M **must be the midpoints of **Segments
DE**, **DF**, and **EF**, respectively. By verifying each
of these midpoints, the proof will be complete.

In order for **L** to be the midpoint
of **Segement DF**, **Segements DL** and **LF** must
be congruent. One way to verify this congruence is by looking
at **Triangles ALF** and **ELD**.

(1) **Angle ALF** is congruent to
**Angle ELD** because they are verticle angles.

(2) Since **D** and **E** are midpoints
of **Segments AB** and **BC**, respectively, **Segment
DE **is a midsegment of **Triangle ABC**. Therefore, the
length of **Segment DE** is half the length of **Segment AC**.
Since **F** is the midpoint of **Segment AC**, **Segment
AF** is half the length of **Segment AC**. Therefore, **Segements
AF** and **DE** must be congruent.

(3) The four smaller triangles formed
by the construction of a medial triangle are each similar to the
original triangle. Thus **Angles LFA** and **LDE** are congruent
as they are corresponding parts of similar triangles.

From (1), (2), and (3), **Triangles
ALF** and **ELD** are congruent by Angle-Angle-Side. Thus,
**Segments DL** and **LF** are congruent (since corresponding
parts of congruent triangles are congruent) and **L** is the
midpoint of **Segment DF**.

Through a similar technique, it can be
shown that **K** is the midpoint of **Segment DE** and **M**
is the midpoint of **Segment EF**. Thus, **Segments DM**,
**EL**, and **FK** are medians of **Triangle DEF** and
**N** is the centroid. QED : )

**Conjecture #2: The circumcenter
of a triangle is the orthocenter of the medial triangle.**

To begin considering this hypothesis, examine the figures below.

Notice that this conjecture appears to be true for both triangles, even when the point in question lies outside of both triangles. A proof of Conjecture #2 can be found below.

**Given:**
**Triangle DEF** is the medial triangle of **Triangle ABC**;
**Triangle ABC** has circumcenter at point **M**,

**Prove: M** is the orthocenter of **Triangle DEF**.

Consider the figure below:

In order for **M** to be the orthocenter
of **Triangle DEF**, **Segements RF**, **PD**, and **QE**
must be altitudes. This means that **Lines RF**, **PD**,
and **QE** must be perpendicular to **Segments DE**, **EF**,
and **DF**, respectively.

This proof will begin by showing that
**Line RF** is perpendicular to **Segment DE**. Since **M**
is the circumcenter of **Triangle ABC**, **Line RF** is
the perpendicular bisector of **Segment AC**. Since **Segment
DE** is a midsegment of **Triangle ABC**, it is parallel
to **Segment AC**. If a line ** m** is perpendicular
to any segment

Through a similar argument, it can be
shown that **Line PD** is perpendicular to **Segment EF**
and **Line QE** is perpendicular to **Segment DF**. Thus,
**Lines** **RF**, **PD**, and **QE** are perpendicular
to the sides of **Triangle DEF**, making **Segments RF**,
**PD**, and **QE** altitudes and, as a result, **M**
the orthocenter. QED : )

**Further Conjectures**

**Conjecture #3: The incenter
of a triangle, the incenter of its medial triangle and the centroid
of the triangles are collinear.**

**Conjecture #4: From Conjecture
#3, the distance from the incenter of the triangle to the centroid
is twice the distance from the incenter of its medial triangle
to the centroid.**

**Conjecture #5: The orthocenter
of a triangle, the orthocenter of its medial triangle and the
centroid of the triangles are collinear.**

**Conjecture #6: From Conjecture
#5, the distance from the orthocenter of the triangle to the centroid
is twice the distance from the orthocenter of its medial triangle
to the centroid.**

**Conjecture #7: The circumcenter
of a triangle, the circumcenter of its medial triangle and the
centroid of the triangles are collinear.**

**Conjecture #8: From Conjecture
#7, the distance from the circumcenter of the triangle to the
centroid is twice the distance from the circumcenter of its medial
triangle to the centroid.**

The proof of these conjectures are left to you. The use of coordinate proofs may be the best approach, as slope and the distance can be expressed in terms of coordinates. Rachael Brown, another doctoral student at UGa, has a page dealing with coordinate proofs and the centers of triangles that may a helpful reference. To view her page, Click Here.