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...
Post a Comment

Menelaus's Theorem


If a transversal cuts the sides $BC, CA, AB$ (extended if necessary) of a triangle $ABC$ at $P, Q, R$ respectively then, $ \frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = -1$, where all segments in the formula are directed segments.
Proof:

Through $A$, draw a line parallel to $QP$ to intersect $BC$ at $K$.

In $ \bigtriangleup RBP$ and $ \bigtriangleup ABK$,

$ \angle RPB = \angle AKB$   (corresponding angles)
$ \angle PRB = \angle KAB$   (corresponding angles)
Therefore, $ \bigtriangleup RBP$ $\sim$ $ \bigtriangleup ABK$ by $AA$ Similarity.
$ \Rightarrow $ $ \frac{AR}{RB} = \frac{KP}{PB}$              ...$(1)$
In $ \bigtriangleup QCP$ and $ \bigtriangleup ACK$,

$ \angle CPQ = \angle CKA$   (corresponding angles)
$ \angle CQP = \angle CAK$   (corresponding angles)
Therefore, $ \bigtriangleup QCP$ $\sim$ $ \bigtriangleup ACK$ by $AA$ Similarity.
$ \Rightarrow $ $ \frac{QC}{QA} = \frac{PC}{PK}$              ...$(2)$

 Multiplying eq.$(1)$ and eq.$(2)$, we get

 $ \frac{AR}{RB} \cdot \frac{QC}{QA} =  -\frac{PC}{PB}$ 

$ \Rightarrow $ $ \frac{AR}{RB} \cdot \frac{QC}{QA} \cdot \frac{PB}{PC}=  -1$
$or$
$ \frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB}=  -1$

Comments

Post a Comment

Popular posts from this blog

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...
Read more

Irrational Numbers

A number which cannot be written in the form $ \cfrac {p}{q}$  where $p$ and $q$ are integers and $ q $  $ \neq 0$. Example: $ \sqrt {2}$, $ \sqrt {3}$, $ \sqrt {6}$, $ \sqrt {7}$, $ \sqrt {8}$, $ \sqrt {10}$ Theorem: If $p$ divides $x^3$, then $p$ divides $x$, where $x$ is a positive integer and $p$ is a prime number.  Proof :  Let $x = p_{1} p_{2}...p_{n}$ where $p_{1}, p_{2}, p_{3}.....p_{n}$  are primes, not necessarily distinct.  $ \rightarrow x^3 = p_{1}^3 p_{2}^3.....p_{n}^3$ Given that $p$ divides $x^3$ By fundamental theorem, $p$ is one of the primes of $x^3$.  By the uniqueness of fundamental theorem, the distinct primes of $x^3$ are same as the distinct primes of $x$.  $ \rightarrow p$ divides $x$ Similarly if $p$ divides $x^2$, then $p$ divides $x$, where $p$ is a prime number and $x$ is a positive integer. Example: Prove that  $ \sqrt {2}$ is irrational.  Solution : Let us assume that  $ \sqrt ...
Read more

Incenter/Excenter Lemma

Image
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...
Read more