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.

Theorem

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.

Proofs

Proof 1

Let $ABC$ be an acute triangle and $C(O,R)$ be its circumcircle. Suppose $BD$ the diameter of $C$ and bring the chord $DC$. Then

$\widehat{BAC}=\widehat{BDC}$ and in the right triangle $BCD$,

$sin\widehat{BDC}=\frac{BC}{BD}=\frac{a}{2R}$

So  $sin\widehat{BAC}=sin\widehat{BDC}=\frac{a}{2R}\Leftrightarrow\frac{a}{sinA}=2R$

If $ABC$ was was an obtuse one, we would follow the same way with the difference that $\widehat{BAC}+\widehat{BDC}=180$ and still $sin\widehat{BAC} =sin\widehat{BDC}$

If, finally, $ABC$ 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$E$ $ABC$ is given by $E=\frac{1}{2}bcsin\widehat{BAC}$ and $E=\frac{abc}{4R}$ (see the topic about Area’s Formulas which i have not yet publiched :)) where $a,b,c$ and $R$ have their common meaning.

So $E=\frac{1}{2}bcsin\widehat{A}=\frac{abc}{4R}\Leftrightarrow sin\widehat{A}=\frac{a}{2R}\Leftrightarrow\frac{a}{sinA}= 2R$ QED

Results

• $\frac{a}{b}=\frac{sinA}{sinB}$ a very importand result.
• $sin(B+C)=sinBcosC + sinCcosB \Leftrightarrow$$\Leftrightarrow sinA2R=2RsinBcosC + 2RsinCcosB$$\Leftrightarrow a=bcosC + ccosB$
• Something obvious, $a>b \Leftrightarrow sinA>sinB$

Exercises

1. 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.
2. If A, B, C, D are concyclic and line AB cuts CD at E, prove that $\frac{AC}{BC} \cdot \frac{AD}{BD} = \frac{AE}{BE}$. (You will need the power of a point theorem)