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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Simson Line

In geometry, given a triangle $ABC$ and a point $P$ on its circumcircle, the three closest points to $P$ on lines $AB$, $AC$, and $BC$ are collinear. The line through these points is the Simson line of $P$, named for Robert Simson.

Prove that the feet of perpendiculars drawn from a point on the circumcircle of a triangle on the sides are collinear.

Solution:
Let $D, E, F$ be the feet of perpendiculars drawn from a point $P$ on the circumcircle of $ \bigtriangleup ABC$ on the sides $BC, CA, AB$ respectively.


We shall prove that the points $D, E, F$ are collinear by showing that
$ \angle PED + \angle PEF = 180 ^\circ $

We start by noting that
$ \angle PEA + \angle PFA = 180 ^\circ $
Therefore, the points $P, E, A, F$ are concyclic.
Consequently,
$ \angle PEF + \angle PAF$                 (angles in the same segment)   $...........(1) $
Since $ \angle PEC = \angle PDC = 90^\circ $, therefore $P, E, D, C$ are concyclic
Therefore $ \angle PED + \angle PCD = 180 ^\circ ............(2)$

Since $P, A, B, C$ are concylic,
therefore $ \angle PAF = \angle PCB............(3)$

From eq.$(1)$ and eq $(3)$,
$ \angle PEF = \angle PCD............(4)$

From eq.$(2)$ and eq $(4)$,
$ \angle PED + \angle PEF = 2 $ right angles.

Hence $D, E, F$ are collinear.




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Equilateral and Equiangular Polygons

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