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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

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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...
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Ceva's Theorem

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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]}$        ...
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Stewart's Theorem

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Stewart's Theorem states:  $$t^2=\frac{b^2u+c^2v}{a}-uv$$ The theorem can also be written as: $$b^2u+c^2v=(a)(uv+t^2)$$ Proof: In $\triangle AED$, by Pythagorean theorem, we have $x^2+h^2=t^2$                                                                                                                                    (1) Similarly, in $\triangle AEC$ $h^2+(v-x)^2=b^2 \Rightarrow h^2u+uv^2-2uvx+ux^2=b^2u$                                               (2) In $\triangle AEB$, we use Pythagorean theorem again and get ...
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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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More Coordinate Geometry

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We have looked at some basics of coordinate geometry. Today, let's look at some intermediate applications of this topic. 1. Shoelace Theorem The Shoelace Theorem is an algorithm used to find the area of a polygon in the coordinate plane when the coordinates are known. If the coordinates are $(x_1, y_1), (x_2, y_2) ... (x_n, y_n), $ $\text{Area}=\frac{1}{2}\Big[(x_1y_2 - x_2y_1)+(x_2y_3 - x_3y_2) + ... + (x_ny_1 - x_1y_n)\Big].$ 2. Centroid The centroid is the point where the three medians intersect. It is also sometimes called the center of gravity for the triangle. Note: A median of a triangle is the line segment joining the vertex to the midpoint of the opposite side. If $(x_1,y_1), (x_2, y_2),$ and $(x_3,y_3)$ are the vertices of a triangle, then the coordinates of its centroid are $$\Bigg(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \Bigg).$$ 3. Incenter The incenter is the point where the three angle bisectors intersect. Note: An angle bisector of a triangle...
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Incenter/Excenter Lemma

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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 = \angl...
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