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The Extended Law of Sines

November 11, 2009

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.


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,


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 triangleE 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


  • \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


  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)


From → Theorems

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