**Assignment 4**

**I chose to write up activity #6:**

I first created a GSP file that contained a triangle, its medial triangle, and the points G, H, C, and I of each of the two triangles. This file can be downloaded here. After playing with this sketch, I noticed several interesting things about these points, which I prove below.

**1. The original triangle's circumcenter is equal to the medial triangle's
orthocenter.**

First, notice that **FE** is parallel to **CA**.
We see why this is true by noticing that angle** FEB** is congruent
to angle **DAE**, since all four of the smaller triangles are congruent.
By the same argument, we can show that **DE **is parallel to **CB**,
and that **DF** is parallel to **AB**. This means
that that the perpendicular bisector of **BC** is the same line
as the altitude of the medial triangle through **F**. Similarly,
the perpendicular bisectors of **AB** and **AC** are
the same as the altitudes of the medial triangle through **E**
and **D**, respectively. This proves our hypothesis.

**2. The original triangle's centroid is equal to the medial
triangle's centroid.**

I claim that the median **BD** of the original triangle
intersects **FE** at the midpoint** H** of **FE**,
therefore **HD **is a median of the medial triangle. This is true,
because since **FE** is parallel to **CA**, triangle
**HEB** is similar to triangle **DAB**. Since **EB**
is half of **AB**, **HE** must be half of **DA**,
so ** H** must be the midpoint of **FE**. By the same
logic, **FG** and **IE** are also medians of the medial
triangle, and so the centroids of the two triangles are equal.

**3. The Euler Segment of the medial triangle is half as
long as the Euler Segment of the original triangle.**

In the following diagram, I have labeled the original triangle's
centroid, orthocenter, and circumcenter as **G**, **H**,
and **C**. Similarly, the medial triangle's centroid, orthocenter,
and circumcenter are labeled **G'**, **H'**, and **C'**.
**G **lies on **CH**, which is the original triangle's
Euler Segment, such that **length(GH) = 2*length(GC)**. Similarly
**G'** lies on **C'H'** such that **length(G'H')
= 2*length(G'C')**.

This means that **length(C'H') = 3*length(C'G')**
and **length(HC) = 3*length(G'H')**. But since **length(G'H')
= 2*length(C'G')**, then **length(HC) = 6*length(C'G')**.
Therefore the length of the medial triangle's Euler Segment (**C'H'**)
divided by the length of the original triangle's Euler Segment (**CH**)
is **[3*length(C'G')] / [6*length(C'G')****] = 1/2**.