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 . The definitions of *middle points, medians, heights, circumcircles* are taken for granted. Thus we start from a very simple property:

**Lemma 1 – The most useful and fundamental lemma in plane geometry**

*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 be the middle points of sides respectively. Suppose that segments meet at and that cuts in . We have to prove that . We take a point in the extension of segment so as . Applying Lemma 1 to triangles and we get that is parallel to and to . As a result, is a parallelogram and thus its diagonals, 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 .

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.

Enjoy!

]]>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,

*kalagz*

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

or

“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”

**Proofs **

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

As in the figure,

since their sum is

We will use that and the formulas

,

(*Product to Sum formula),*

*(Sum to Product formula)*

Applying the extended law of sines to triangles ADB, DBC respectively and the* *Sum to Product formula we obtain:

(1)

Similarly, (2)

So

But

(3)

And

Furthermore

(4)

And in the same way (5)

Combining the results 3, 4 and 5 and using again the Law of Sines:

And the proof is completed.

**Proof 2 of the direct theorem** (With Areas)

We know that in a convex quadrilateral ABCD, its area can be given by:

where P is the intersection of AC and BD and from now on S(x) will present the area of polygon “x”.

We will find a relation between and (ABCD).

Take E, concyclic with points A,B,C,D such that . So implying that where is the arc “x”.

It is and using this,

(since )

.

So we conclude that

**Proof of the converse of the theorem **(By Inversion)

For the prove of the converse of Ptolemy’s 1st Theorem we will work with inversion. Suppose that (1). We will prove that ABCD is inscribed.

Consider the inversion with center A and any radius r (suppose r<min{AD,AB} as in the figure). Let the images of B,C,D be B’,C’,D’ respectively. By inversion, triangles ABC-AC’B’, ACD-AD’C’ and ABD-AD’B’ are similar which implies that , , . Introdusing these in (1) we get

implying that B’, C’ and D’ are collinear. By inversion, this means that points B, C and D are concyclic, so the proof is completed.

**Ptolemy’s Second Theorem**

If a quadrilateral ABCD is cyclic then, for the diagonals AC and BD, it is true that

**Proof**

In every triangle with sides and the radius of the circumcircle of the triangle and the height to side a respectively, it is known that . So, according to the figure,

(1)

And similarly (2). Dividing 1 with 2 we get the desired relation.

**Notes**

– By Ptolemy’s theorem, we can produce Ptolemy’s Inequality:

For every quadrilateral ABCD,

with equality if and only if ABCD is cyclic.

- Proof of direct Pythagorean Theorem. Let ABDC be a rectangle, obviously a cyclic quadrilateral. By Ptolemy’s we obtain

- Let ABC be an equilateral triangle of side a and P a point in its plane. Then P belongs to the circumcircle of the triangle if and only if PB=PA+PC. We have equivalently ABCP is inscribed.
**Mahavira’s Theorem**In a cyclic quadrilaral with sides a, b, c, d and diagonals m, n (as in the figure) the length of the diagonals are given by and

**Exercises**

**Historical Facts**

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

**Theorem**

In any triangle the cevians are concurrent if and only if

(simple form)

or

(trigonometric form)

**Proofs**

**Proof 1**

First we prove that we have concurrency if and only if . 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

(1) (2) and (3) Multipling relations 1, 2 and 3 we get the desired.

Now suppose that . 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 . Dividing the last two relations we get that and since D and D’ both lie in segment BC, we get , 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 :

. Similarly we get and . Multipling these relations we get the desired .

For the converse we work like the first proof to get .

*Proof of the trigonometric form of Ceva’s theorem*

If AD, BE, CF are concurrent, say at P, by the Law of Sines at triangles ABP, BCP, CPA we get (1) (2) (3) and multipling these we get the desired.

Finally, we have to show that if then AD, BE, CF concur.

By the Law of Sines at triangles ABD and ACD we get (1) and (2). But since they are complementa angles. So, dividing relations 1 and 2, . Similarly, and . Multipling the last three relations and using the hypothesis we get which is enough for what we need since we have prove it earlier.

**Notes**

– The points D, E, F can be on the extensions of the sides of the triangle

**Results**

- The medians of a triangle concur at the centroid. Simply aplly Ceva’s theorem using where are the midpoints of respectively.

- The internal angle bisectors at the incenter. By the intarnal angle bisector theorem, if the bisector of then etc. Multiply the 3 equations and QED.
- The three altitudes of every triangle concur at the orthocenter of the triangle. If are the projections of at the opposite sides, use the trigonometric form of Ceva with and
- The cevians joining the points of tangency of the incircle with the sides of a triangle concur at the
**Gergonne****point**of the triangle. If the desired points, then so by Ceva’s simple form the conclusion is obvious. - Similarly the cevians that connect the points of tangency of the 3 excircles with the corresponding sides of a tringle also concur at
**Nagel point.***(The proof comes soon)*

**Exercises**

- Let the ceavians AD, BE, CF be concurrent in a triangle ABC. Prove that FE is parallel to BC if and only if AD is the bisector of BC
- In a triangle ABC the median AM meets the internal bisector BN at P. Let Q be the intersection of AB and CP. Prove that BNQ is an isosceles triangle.
- Let ABC be a triangle and let D, E, F be the points on sides BC, CA, AB respectively such that the cevians AD, BE, CF are concurrent. Let M, N, P be points on EF, FD, DE respectively. Prove that lines AM, BN, CP concur if and only if lines DM, EN, FP concur.
- (*)In a triangle ABC points D, E, F are on sides BC, CA, AB so that cevians AD, BE, CF concur. Let the parallels to the line AB from points E and Q meet the lines DF and EF at Q and R respectively. Prove that lines CF, DE, QR are concurrent.

$latex \frac{BD}{DC}= * *\frac{(BDA)}{(CDA)}= \frac{(BDP)}{(DCP)}= $ . Similarly we get and $latex \frac{AF}{FB}= \frac{(CAP)}{(

C

BP)} $. Multipling these relations we get the desired

For the converse we work like the first prove to get

**Theorem**

In any triangle with angles and sides respectively the following is true

where is the circumradii of the triangle.

**Proofs**

**Proof 1 **

Let be an acute triangle and be its circumcircle. Suppose the diameter of and bring the chord . Then

and in the right triangle ,

So

** **

If was was an obtuse one, we would follow the same way with the difference that and still

If, finally, was a right triangle, then the proof would be obvious. *(I leave this proof for the reader :))*

*Proof 2 **(Using Area formulas)*

The area of every triangle is given by and (see the topic about Area’s Formulas which i have not yet publiched :)) where and have their common meaning.

So QED

**Results**

- a very importand result.
- Something obvious,

**Exercises**

- If in a triangle ABC, A=45 and b=R=1, where R is the radious of the circumcircle of the triangle, find side c.
- If A, B, C, D are concyclic and line AB cuts CD at E, prove that . (You will need the power of a point theorem)

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 (www.Mathlinks.ro).That’s all i have to say for introduction. Thanks for reading this and using my collection.

Now it is time for Geometry…

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