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From the medians to the Theorem of Feuerbach

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.

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.

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.


Back @ Work – Back to Geometry !

Hello again

This is something like re-introduction post. Lot of things have happened since my last posting. Participation in the International Mathematical Olympiad, Medals in many kinds of math olympiads, entrance in university, IMF in Greece, War in Syria. Of course the last two have nothing to do with this blog.

The important thing is that I am back. This means that from now on, this blog will be very active as regard to new posts and changes. So I hope it will also be active from your point of view, the viewers.

I would also like to mention my motivation for fast editing new posts. This blog had some unexpected visitors the past two months. When I noticed this in the stats, I was very suprised and happy. I must admit that I felt so happy as I was when I got informed for my bronze medal In the 52th IMO!

Anyway, I stop speaking and I start Geometry.


Blog inactive due to exams obligations

Dear Reader
(if you exist…)

Recently, i succeded in the exams for the Greek team for Mathematical Olympiads. As a result i will participate (as GRE5) in the 28th Balkan Mathematical Olympiad [4-8 May 2011] and in the 52nd International Mathematical Olympiad [16-24 July 2011] (this is not sure but very possible). In the meanwhile, I will be giving school exams for university introduction. In addition, in August i “will have to take” some vacations!

So i will not be able to post something new before September. But i promise a dynamic comeback in Autumn, with many posts, difficult exercises, rich material for Olympiad Geometry and many PDF files with blog’s context.

Best regards,

Ptolemy’s Theorems


Found by Ptolemy of Alexandria (or Claudius Ptolemaeus) (90-168 AD).

Ptolemy’s First Theorem

A quadrilateral ABCD is cyclic (inscribed) if and only if

AB\cdot CD + AD\cdot BC=AC\cdot BD


“A quadrilateral is inscribed in a circle if and only if the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals”


Proof 1 of the direct theorem(Using a little Trigonometry :))

As in the figure, a=a_1, b=b_1, c=c_1, d=d_1,

\sin(a+b)= \sin(d+c_1),

\sin(c+a_1)= \sin(b_1+d_1) since their sum is \pi

We will use that \cos X=- \cos(180^\circ-X) and the formulas

\cos(90^\circ-X)= \sin X ,

2 \sin X\cdot \sin Y= \cos(X-Y)- \cos(X+Y) (Product to Sum formula),

\cos X+ \cos Y=2\cdot \cos \frac{X+Y}{2}\cdot \cos \frac{X-Y}{2} (Sum to Product formula)

Read more…

Ceva’s Theorem

Cevian is called every segment that connects a vertice of a triangle with the opposite side. Three or more lines are concurrent when all pass through a common point, so we are talking about concurence.

Historical Facts

This theorem was published from Giovanni Ceva (in honour of whom the segments before are named ) in 1678.


In any triangle ABC the cevians AD, BE, CF are concurrent if and only if

\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1 (simple form)


\frac{sinBAD}{sinDAC}\cdot\frac{sinABE}{sinEBC}\cdot\frac{sinBCF}{sinFCA}=1 (trigonometric form)


Proof 1

First we prove that we have concurrency if and only if \frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1 . Assume that AD, BE, CF meet at point P. We extend BE and CF until they meet the parallel to BC from A at G and H respectively.

Then, we have the relations

\frac{AF}{FB}= \frac{AH}{BC} (1) \frac{CE}{EA}= \frac{BC}{AG} (2) and \frac{BD}{DC}= \frac{BD}{DP} \cdot \frac{DP}{DC}= \frac{AG}{PA} \cdot \frac{PA}{AH} (3) Multipling relations 1, 2 and 3 we get the desired.

Now suppose that \frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1 . We have to prove that AD, BE, CF are concurrent. Let BE and CF meet at P and suppose AP cuts BC at D’.

We get by what we have proved so far that  \frac{BD'}{D'C}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1 . Dividing the last two relations we get that \frac{BD}{DC} = \frac{BD'}{D'C} and since D and D’ both lie in segment BC, we get D \equiv D' , QED

Proof 2 (Using areas)

We use the fact that the areas of triangles with equal altitudes are proportional to the bases of the triangles. Let AD, BE, CF concur at P. We have :

\frac{BD}{DC}= \frac{(BDA)}{(CDA)}= \frac{(BDP)}{(DCP)} \Leftrightarrow \frac{BD}{DC}= \frac{(BDA)-(BDP)}{(CDA)-(CDP)}= \frac{(APB)}{(APC)} . Similarly we get \frac{CE}{CA}= \frac{(BCP)}{(BAP)} and \frac{AF}{FB}= \frac{(CAP)}{(CBP)} . Multipling these relations we get the desired \frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1 .

For the converse we work like the first proof to get D \equiv D' .

Read more…

The Extended Law of Sines

Well, I start the collection with one of the most importand theorems in Geometry, the sines law for every triangle. I think this blog must start with this basic theorem because many others are proved usind this.


In any triangle ABC with angles A,B,C and sides a,b,c respectively the following is true

\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=  2R

where R is the circumradii of the triangle.

Read more…

Welcome to my blog, a collection of theorems in Plain Euclidean Geometry!(Intro)

The purpose of this blog is to collect as many as possible theorems with proofs and applications in one site. I’m just a school-boy who takes part in mathematical competitions and loves geometry. My “addiction” to Classical Euclidean Geometry made me create this blog. One more reason is to have an online collection with many theorems organized and “well-given” with some applications and exercises.
Every theorem in this blog will be consisted of some historical facts (if i know any) the theorem itself ofcourse, many proofs, some remarks, other results(findings) and 3-4 exercises in increasing difficulty level taken from school books to Highly level Maths Olympiads.
This collection refers to every student who wants to learn a bit more about Classic Euclidean Geometry, the guy who participates in math contests in every level and to everyone who loves solving geometry problems. Of course some of you may not need this but as i said, this is just a collection which aims to have complete posts about every theorem.
You can help this work become better! Something great with blogs is that you can post a comment. So i would like every reader to make a comment if he has to suggest any improvement or a question about something posted.
In this point i would really like to thank Art Of Problem Solving society (’s all i have to say for introduction. Thanks for reading this and using my collection.
Now it is time for Geometry…