Comparison: Solution Of Triangles In Trigonomemetry: Napoleon

When I was given to write on the topics of Napoleon’s Theorem, the first thing that struck my mind was that it was somehow related to the French leader, Napoleon Bonaparte. But then a thought struck me: Napoleon was supposed to good at only politics and the art of warfare. Mathematics was never related to him. On surfing the internet to learn about the theorem, I came to know that this theorem was in fact named after the same Napoleon as he was good at Maths too (other than waging wars and killing people). The theorem was discovered in his ruling period and was named so in his honour.
Now moving on the towards the theorem part, it is a very observation made by some of the brightest minds of that time. Sorry to disappoint you, but the theorem
First I will state the Cosine Law which deals with finding the side of a triangle if the sides and angles are given. Consider the following figure- In the above given triangle ABC, sides AB, BC, and CA are represented by the small letters c, a and b. These are used as it is a rule to denote a side opposite to a particular angle in a triangle with the same letter in lower case.
Thus the formula is given by – a2 = b2 + c2 – 2bc(CosA)
Similarly, other sides can also be found
If we want to find height CM, it can be either aSinB or bSinA.
Thus Area = acSinB/2 or bcSinA/2 or abSinc/2.

Now coming back to Napoleon’s Theorem, consider the following figure where ABC is the original triangle and triangles FAB, EAC and CDH have been constructed on the sides of the triangle which have G, I and H as their respective centres. These centres are joined and a triangle GHI is obtained. Now all we have to do is to prove that the length of each side of triangle GHI is equal to s (which is considered to be the length GI).To help me in my proof, I gave taken length AG as t and length AI as u. Now since G, I and H are the respective centres of each equilateral triangle, they act as both incentres too. Thus the length t and u bisect the angles FAB and EAC into two 30 degree angles. Since we know the measure of angle GAB and IAC, we can now use the Cosine Law for the triangle AIG. It will be given as – s2 = t2 + u2 – 2tu(Cos(A+60))
On using formula Cos(A+B)= CosA CosB – SinA SinB, we get – s2 = t2 + u2 – 2tu(CosA Cos60 – SinA Sin60) s2 = t2 + u2 – tu(CosA) + tu√3(SinA)
Now we find t and u in terms of c and