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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Incenter/Excenter Lemma

Given $\triangle ABC$ with incenter $I$, extend $AI$ to meet the circumcircle of $\triangle ABC$ at $L$. Then, reflect $I$ over $L$, and name this point $I_a$. Then:

  $(i)$ $BICI_a$ is a cyclic quadrilateral with point $L$ as the center of the circumscribed circle.
 $(ii)$ $BI_a$ and $CI_a$ bisect the exterior angles of $\angle B$ and $\angle C$, respectively. 

Proof
$(i)$ Notice that the the question is equivalent to proving the distance from $L$ to points $B, I, C, I_a$ are equal. In other words, we need to show that 
$$LB = LI = LC = LI_a$$

 Firstly, it is obvious that $LI = LI_a$ because by definition, $I_a$ is the reflection of $I$ over $L$, and reflections preserve the length of the segment. 

 Now, we are left with proving that $LB=LI$, because similar calculations will provide $LI=LC$. 

 We see that most of our given information involves angles, so we focus on proving $\angle LBI = \angle LIB$. Firstly, 
$$\angle LBC = \angle LAC = \cfrac{A}{2}$$.
Then, note that $BI$ is an angle bisector, so
$$\angle IBL = \angle LBC + \angle IBC = \cfrac{A}{2} + \cfrac{B}{2} = \cfrac{A+B}{2}$$

 We are nearing the finish line, with the only thing remaining is to prove $\angle LIB = \angle IBL$. To do this, we look at $\triangle AIB$, because $\angle LIB$ is an exterior angle of the triangle. Using the exterior angle theorem, we see that 
$$\angle LIB = \angle AIB + \angle IBA = \cfrac{A}{2} + \cfrac{B}{2} = \cfrac{A+B}{2} = \angle LBI \text{ } \Box$$

 $(ii)$ For the second part of this proof, we need to prove $\angle CBI_a = \cfrac{1}{2} \cdot \angle CBD$. This is equivalent to showing that $\angle CBI_a = \cfrac{180-B}{2}$ because $\angle CBD = 180 - B$.

 In the previous part, we proved that $II_a$ is a diameter of the circle, so $\angle IBI_a = 90$. We know that $\angle IBC = \cfrac{B}{2}$, so 
$$\angle CBI_a = \angle IBI_a - \angle IBC = 90 - \cfrac{B}{2} = \cfrac{180-B}{2}$$.
Similar logic provides $CI_a$ is the bisector of the exterior angle at $C$, and we are done. 

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Equilateral and Equiangular Polygons

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