Equilateral and Equiangular Polygons

A polygon is a 2 dimensional geometric figure bound with straight sides. A polygon is called Equilateral if all of its sides are congruent. Common examples of equilateral polygons are a rhombus and regular polygons such as equilateral triangles and squares.  Now, a polygon is equiangular if all of its internal angles are congruent.  Some important facts to consider The only equiangular triangle is the equilateral triangle If P is an equilateral polygon that has more than three sides, it does not have to be equiangular. A rhombus with no right angle is an example of an equilateral but non-equiangular polygon.  Rectangles, including squares, are the only equiangular quadrilaterals Equiangular polygon theorem. Each angle of an equiangular n-gon is  $$\Bigg(\frac{n-2}{n}\Bigg)180^{\circ} = 180^{\circ} -   \frac{360^{\circ}}{n} $$ Viviani's theorem   Vincenzo Viviani (1622 – 1703) was a famous Italian mathematician. With his exceptional intelligence in math...
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Ceva's Theorem

If the lines joining the vertices $A, B, C$ of a triangle $ABC$ to any point $P$ in their plane meet the opposite sides in $D, E, F$ then $ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=  1$
Proof: 
In $ \bigtriangleup AFP$ and $ \bigtriangleup FBP$, the height of the altitude is the same if we drop a perpendicular from $P$ to $AF$ and $FB$.
Therefore,
$ \frac{AF}{FB} =  \frac{[AFP]}{[FBP]}$                           ...(1)

Similarly, in $ \bigtriangleup AFC$ and $ \bigtriangleup FBC$, the height of the altitude is same if we drop a perpendicular from $C$ to $AF$ and $FB$.
Therefore,
$ \frac{AF}{FB} =  \frac{[AFC]}{[FBC]}$                           ...(2)

By subtracting the triangle areas of the eq.(2) with eq.(1), we get
$ \frac{AF}{FB} =  \frac{[APC]}{[BPC]}$                           ...(3)

Similarly,
$ \frac{BD}{DC} =  \frac{[APB]}{[APC]}$                   ...(4)   
and
$ \frac{CE}{EA} =  \frac{[BPC]}{[APB]}$                    ...(5)

Multiplying eq. (3) and (4) and (5), we get

$ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = \frac{[APC]}{[BPC]} \cdot \frac{[APB]}{[APC]} \cdot \frac{[BPC]}{[APB]} = 1$





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Equilateral and Equiangular Polygons

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