An easy but also deep journey on the geometry of a triangle and its exciting topics.
[Under construction…]
On this essay-course we begin studying some basic properties and gradually we touch some of the most interesting and “advanced” facts of the most simple plot in the Euclidean Geometry, the triangle. A skllied problem solver should be familiar with the biggest part of what we are going to mention. Nevertheless, we wish to show how simple the connection is between basic principles and advanced properties.

We did the same on the exercises in the end of this post. The first are very simple but we hope that the last will be tough even for skilled problem solvers.

All we need is an acute-angled triangle $ABC$. The definitions of middle points, medians, heights, circumcircles are taken for granted. Thus we start from a very simple property:

Lemma 1The most useful and fundamental lemma in plane geometry
The line segment joining the middle points of any two sides of a triangle is parallel to and equals half of the third side.

Proof
We skip it. Don’t be afraid to prove this on your own, think as an entrance test for the rest. What you need is a parallelogram. Unfortunately, if you can’t prove this, you will meet troubles with what follows.

We move to our first theorem

Theorem 1
The medians of a trianlge are concurrent; all pass through the same point.

Proof
Let $L, N$ be the middle points of sides $AB, AC$ respectively. Suppose that segments $BN, CL$ meet at $G$ and that $AG$ cuts $BC$ in $M$. We have to prove that $BM=MC$. We take a point $Z$ in the extension of segment $AG$ so as $AG=GZ$. Applying Lemma 1 to triangles $ABZ$ and $ACZ$ we get that $LG$ is parallel to $BZ$ and $GN$ to $ZC$. As a result, $BGCZ$ is a parallelogram and thus its diagonals, $GZ,BC$ are bisected. The proof is now completed.

Definition 1
The point of intersection of the three medians of any triangle is called Centroid. From now on, we will represent this point with $G$.