lunes, 6 de julio de 2020

Generalization of two formulae and an alternative proof of Bretschneider's formula

"If we do not succeed in solving a mathematical problem, the reason frequently consists in our failure to recognize the more general standpoint from which the problem before us appears only as a single link in a chain of related problems. After finding this standpoint, not only is this problem frequently more accessible to our investigation, but at the same time we come into possession of a method which is applicable also to related problems."  David Hilbert
 

The following formulae generalize $(1)$ in my previous post Killing three birds with one stone. For implications in a triangle see also Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle

Here, $a$, $b$, $c$, $d$ are the sides of a general convex quadrilateral, $s$ is the semiperimeter, and $\alpha$ and $\gamma$ are two opposite angles. Then



$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)-bc\cos^2{\frac{\gamma}{2}}}{ad}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)-bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{1}$$

Proof. By the Law of Cosines,

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{\gamma}\tag{2}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2+2bc\cos{\gamma}}{2ad}$. Now, making use of the half angle formula for cosine,

$$\begin{align*} \cos^2{\frac{\alpha}{2}}&=\frac{a^2+d^2+2ad-b^2-c^2+2bc\cos{\gamma}}{4ad}\tag{3}\\ &=\frac{a^2+d^2+2ad-b^2-c^2+2bc(1-2\sin^2{\frac{\gamma}{2}})}{4ad}\tag{4}\\&=\frac{(a+d)^2-(b-c)^2-4bc\sin^2{\frac{\gamma}{2}}}{4ad}\tag{5}\\&=\frac{(a+d+b-c)(a+d-b+c)-4bc\sin^2{\frac{\gamma}{2}}}{4ad}\tag{6}\\&=\frac{1}{ad}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)-\frac{bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{7}\\&=\frac{(s-b)(s-c)-bc\sin^2{\frac{\gamma}{2}}}{ad}\tag{8}\end{align*}$$

$\square$

The other formula can be obtained similarly by replacing $\cos^2{\frac{\alpha}{2}}$ by $1 - \sin^2{\frac{\alpha}{2}}$ in $(3)$.

A proof of Bretschneider's formula
The formulae in $(1)$ can be rewritten as follows

$$ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}=(s-a)(s-d)\tag{9}$$

and

$$bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}=(s-b)(s-c)\tag{10}$$

Multiplying $(9)$ and $(10)$ we get

$$\begin{align*}\left(ad\sin^2{\frac{\alpha}{2}}+bc\cos^2{\frac{\gamma}{2}}\right)\left(bc\sin^2{\frac{\gamma}{2}}+ad\cos^2{\frac{\alpha}{2}}\right) &= (s-a)(s-b)(s-c)(s-d)\tag{11}\end{align*}$$

Expanding, factorizing, completing the squares and keeping in mind some well-known trigonometric identities, 

$$\begin{align*}abcd\cos^2\left({\frac{\alpha+\gamma}{2}}\right)+\left(ad\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}+bc\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}\right)^2 &=(s-a)(s-b)(s-c)(s-d)\tag{12}\\abcd\cos^2\left({\frac{\alpha+\gamma}{2}}\right)+\left(\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\gamma}}{2}\right)^2 &=(s-a)(s-b)(s-c)(s-d)\tag{13}
\end{align*}$$

Since the area of $ABCD$ can be expressed as the sum of the areas of $\triangle{ABD}$ and $\triangle{CBD}$, which in turn can be written as $\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\gamma}}{2}$, then we are done.
$\square$

Below you can find a concept map of the identities $(9)$ and $(10)$ so you can see clearly what's going on here (click on the image to have a better view). 


It would be interesting to investigate whether identities $(9)$ and $(10)$ can be generalized to other geometries.

sábado, 4 de julio de 2020

Killing three birds with one stone

In this note we derive Heron's formula, Brahmagupta's formula and the bicentric quadrilateral's area formula, $\sqrt{abcd}$, from two formulae involving sine, cosine, semiperimeter and the side lenghts of a cyclic quadrilateral. 

Let $ABCD$ be a cyclic quadrilateral with $AB=a$, $BC=b$, $CD=c$, $DA=d$ and $s=\frac{a+b+c+d}{2}$. If $\angle{BAD}=\alpha$, then



$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad+bc}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{ad+bc}\tag{1}$$

or 

$$sin^2{\frac{\alpha}{2}}=\frac{(a+b+c-d)(-a+b+c+d)}{4(ad+bc)}\quad and \quad cos^2{\frac{\alpha}{2}}=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\tag{2}$$

Proof. First we will find an expression for $\cos{\alpha}$ in terms of $a$, $b$, $c$ and $d$. Let $\angle{BCD}=\gamma$. By the Law of Cosines and keeping in mind that $\alpha$ and $\gamma$ are supplementary, we have

$$a^2+d^2-2ad\cos{\alpha}=b^2+c^2-2bc\cos{(180^\circ-\alpha)}\tag{3}$$

Yielding $\cos{\alpha}=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}$. Now, making use of the half angle formula for cosine,

$$\begin{align*} \cos^2{\frac{\alpha}{2}}&=\frac{2ad+2bc+a^2+d^2-b^2-c^2}{4(ad+bc)}\tag{4}\\ &=\frac{(a+d)^2-(b-c)^2}{4(ad+bc)}\tag{5}\\&=\frac{(a+b-c+d)(a-b+c+d)}{4(ad+bc)}\tag{6}\\&=\frac{1}{ad+bc}\left(\frac{a+b+c+d}{2}-c\right)\left(\frac{a+b+c+d}{2}-b\right)\tag{7}\\&=\frac{(s-b)(s-c)}{ad+bc}\end{align*}$$

$\square$

The other formulae can be obtained similarly by replacing $\cos^2{\frac{\alpha}{2}}$ by $1 - \sin^2{\frac{\alpha}{2}}$.

A generalization of $(1)$ together with a proof of Bretschneider's formula can be found here.

Remark$(1)$ appears as exercise 400 in V. Panagiotis' 1000 General Trigonometry Exercises, Volume B.

A proof of Heron's Formula
For a triangle, if in $(1)$ we assume $c=0$, then we have

$$\sin^2{\frac{\alpha}{2}}=\frac{(s-a)(s-d)}{ad}\quad and \quad \cos^2{\frac{\alpha}{2}}=\frac{s(s-b)}{ad}\tag{8}$$


Let $\Delta_0$ be the area of $\triangle{ABD}$. Making use of the double-angle identity for sine we have

$$\sin{\alpha}=2\sqrt{\frac{s(s-b)}{ad}}\sqrt{\frac{(s-a)(s-d)}{ad}}=2\frac{\sqrt{s(s-a)(s-b)(s-d)}}{ad}\tag{9}$$

Since $\Delta_0=\frac{ad\sin{\alpha}}{2}$, it follows 

$$\Delta_0=\sqrt{s(s-a)(s-b)(s-d)}\tag{10}$$

$\square$

For more implications in a triangle see here.

A proof of the Brahmagupta's formula
Denote $\Delta_1$ the area of the cyclic quadrilateral, $ABCD$. Then

$$\begin{align*}\Delta_1&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\tag{11}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\tag{12}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\tag{13}\\&=\sqrt{\frac{(s-a)(s-d)}{ad+bc}}\sqrt{\frac{(s-b)(s-c)}{ad+bc}}(ad+bc)\tag{14}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\tag{15}\end{align*}$$

$\square$

A proof of the bicentric quadrilateral's area formula
Since $a+c=b+d$ in a bicentric quadrilateral, the formulae in $(2)$ reduce to 

$$sin^2{\frac{\alpha}{2}}=\frac{bc}{ad+bc}\quad and \quad cos^2{\frac{\alpha}{2}}=\frac{ad}{ad+bc}\tag{16}$$

Assume $ABCD$ is a bicentric quadrilateral and let $\Delta_2$ be its area, then

$$\begin{align*}\Delta_2&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{(180^\circ-\alpha)}}{2}\tag{17}\\&=\frac{ad\sin{\alpha}}{2}+\frac{bc\sin{\alpha}}{2}\tag{18}\\&=\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}(ad+bc)\tag{19}\\&=\sqrt{\frac{bc}{ad+bc}}\sqrt{\frac{ad}{ad+bc}}(ad+bc)\tag{20}\\&=\sqrt{abcd}\tag{21}\end{align*}$$

$\square$

sábado, 20 de junio de 2020

Proofs and applications of two well-known formulae involving sine, cosine and the semiperimeter of a triangle

In Cut-the-knot's Relations between various elements of a triangle, the formulae (a generalization can be found here)

$$\sin^2{\frac{\gamma}{2}} = \frac{(s-a)(s-b)}{ab}\quad and\quad\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}$$ 

are derived using Heron's formulaHere we give an alternative proof without using Heron's Formula and we demonstrate several well-known theorems based on these formulae as a sample of its power. We will be using standard notation: $BC=a$, $AC=b$, $AB=c$, $\Delta$ for the area, $s$ for the semiperimeter, $R$ for the circumradius and $r$ for the inradius. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$.



Notice that $\frac{\cot{\frac{\gamma}{2}}}{s-c} = \frac{1}{r}$. Also, We know $Δ = rs$ and $Δ =\frac {ab\sin{\gamma}}{2}$, hence

$$\frac{\cot{\frac{\gamma}{2}}}{s-c}=\frac{1}{r}=\frac{s}{\Delta}=\frac{2s}{ab\sin{\gamma}}$$

But, $\sin{\gamma} = 2\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}$, so 
 
$$\frac{\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}}\cdot{\sin{\frac{\gamma}{2}}\cos{\frac{\gamma}{2}}} = \frac{s(s-c)}{ab}$$

from which we get $\cos^2{\frac{\gamma}{2}} = \frac{s(s-c)}{ab}$.
 
The other formula can be obtained replacing $\cos^2{\frac{\gamma}{2}}$ by $1 - \sin^2{\frac{\gamma}{2}}$. Indeed, 

$$\begin{aligned} 1-\sin^2{\frac{\gamma}{2}} &= \frac{s(s-c)}{ab} \\ \sin^2{\frac{\gamma}{2}} &= 1-\frac{s(s-c)}{ab} \\  &= \frac{\left(ab-s(s-c)\right)}{ab} \\ &= \frac{\left((y+z)(x+z)-(x+y+z)(z)\right)}{ab}\\ &= \frac{xy}{ab} \\ &=\frac{(s-a)(s-b)}{ab} \end{aligned}$$

$\square$

1. A proof of Heron's Formula
Making use of the formulae proven above and the double angle identity for sine we have

$$\sin{\gamma}=2\sqrt{\frac{s(s-c)}{ab}}\sqrt{\frac{(s-a)(s-b)}{ab}}=2\frac{\sqrt{s(s-a)(s-b)(s-c)}}{ab}$$

Since $\Delta=\frac{ab\sin{\gamma}}{2}$, it follows 

$$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$

$\square$

2. A proof of the Law of Cosines
Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=s(s-c)-(s-a)(s-b)$$
Substituting and multiplying by 4, 
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.


3. Proofs for some trigonometric identities associated to a triangle
a) $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=1$.

As a consequence of the formulae proven at the beginning of the note,

$$\tan{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}, \quad\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\quad and \quad \tan{\frac{\gamma}{2}}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$

 So, by canceling and simplifying you get

$$\begin{aligned}\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\alpha}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}} &=\frac{s-c}{s}+\frac{s-b}{s}+\frac{s-a}{s}\\ &=\frac{z+y+x}{s}\\ &=\frac{s}{s}=1\end{aligned}$$

$\square$

b) $r=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}$.

We make use of the well-known relationship $abc=4R\Delta$ (see here for a proof) and Heron's Formula.

$$\begin{aligned}r&=4R\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-c)}{ac}}\sqrt{\frac{(s-a)(s-b)}{ab}}\\&=4R\sqrt{\frac{(s-a)^2(s-b)^2(s-c)^2}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{a^2b^2c^2}}\\&=4R\sqrt{\frac{\frac{\Delta^4}{s^2}}{16R^2\Delta^2}}\\&=\frac{\Delta}{s}\\&=\frac{rs}{s}\\&=r\end{aligned}$$

$\square$

c) $s=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$.

$$\begin{aligned}s&=4R\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}\\ &=4R\sqrt{\frac{s(s-a)}{bc}}\sqrt{\frac{s(s-b)}{ac}}\sqrt{\frac{s(s-c)}{ab}}\\&=4R\sqrt{\frac{s^2\Delta^2}{a^2b^2c^2}}\\&=4R\frac{s\Delta}{abc}\\&=4R\frac{s\Delta}{4R\Delta}\\&=s\end{aligned}$$

$\square$


Consequently, the following relationship also holds

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{r}{s}$$

or

$$\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\frac{s}{r}$$

d) $\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}$.

To prove the above identity we will show that the right hand side equals $\frac{s}{r}$.

$$\begin{aligned}\frac{s}{r}&= \cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}=\cot{\frac{\alpha}{2}}+\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}\\&=\frac{\sqrt{s(s-a)}}{\sqrt{(s-b)(s-c)}}+\frac{\sqrt{s(s-b)}}{\sqrt{(s-a)(s-c)}}+\frac{\sqrt{s(s-c)}}{\sqrt{(s-a)(s-b)}}\\&=\frac{\Delta(s-a)+\Delta(s-b)+\Delta(s-c)}{(s-a)(s-b)(s-c)}\\&=\frac{\Delta(x+y+z)}{xyz}\\&=\Delta\frac{s^2}{\Delta^2}\\&=\frac{s^2}{rs}\\&=\frac{s}{r}\end{aligned}$$

$\square$


We invite the reader to prove the following identity (possibly new) on their own.
$$\frac{\cot{\frac{\alpha}{2}}\cot{\frac{\beta}{2}}+\cot{\frac{\alpha}{2}}\cot{\frac{\gamma}{2}}+\cot{\frac{\beta}{2}}\cot{\frac{\gamma}{2}}}{(s-a)(s-b)+(s-a)(s-c)+(s-b)(s-c)}=\frac{1}{r^2}$$

4. Proofs for some relationships involving the inradius and exradii

a) This is problem $194$ in Go Geometry.


Proof. Let $\angle{BAC}=\alpha$. If $E'$ is the orthogonal projection of $E_a$ onto $AC$, then $AE'=s$ (well-known). It follows

$$\tan^2{\frac{\alpha}{2}}=\frac{(s-b)(s-c)}{s(s-a)}=\frac{r_a^2}{s^2}$$

Yielding, 

$$r_a=\sqrt{\frac{s(s-b)(s-c)}{s-a}}=\sqrt{\frac{s(s-a)(s-b)(s-c)}{(s-a)^2}}=\frac{\Delta}{s-a}$$

$\square$

Similarly we can get $r_b=\frac{\Delta}{s-b}$ and $r_c=\frac{\Delta}{s-c}$.

sábado, 13 de junio de 2020

Proof Without Words for the Addition Formula for Sine and Subtraction Formula for Cosine


I just hope this is new. What better gift than that on my birthday? :)

$$\sin{(\alpha+\beta)}=\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}\qquad\left(\alpha+\beta<\frac{\pi}{2}\right)$$

$$\cos{(\alpha-\beta)}=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\qquad\left(\alpha-\beta<\frac{\pi}{2}\right)$$



Related material.

Geometric Proof of the Sum Angle Formula (Sine, Cosine)


I wouldn't say the simpler one but at least $\sin(α+β)$ can be obtained almost without words.



$\frac{\cos{\alpha}}{\cos{\beta}}-\frac{\sin{\beta}(\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta})}{\cos{\beta}}=\frac{\cos{\alpha}-\sin^2{\beta}\cos{\alpha}-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=\frac{\cos{\alpha}(1-\sin^2{\beta})-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=$

$\frac{\cos{\alpha}\cos^2{\beta}-\sin{\beta}\sin{\alpha}\cos{\beta}}{\cos{\beta}}=\cos{\alpha}\cos{\beta}-\sin{\beta}\sin{\alpha}=\cos{(\alpha+\beta)}.$

viernes, 5 de junio de 2020

Yet Another Proof of the Law of Cosines

The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. Using standard notation, the law of cosines states

$$c^2=a^2+b^2-2ab\cos{\gamma},$$

where $\gamma$ denotes the angle contained between sides of lengths $a$ and $b$ and opposite the side of length $c$. For the same figure, the other two relations are analogous:

$$a^2=b^2+c^2-2ac\cos{\alpha},$$
$$b^2=a^2+c^2-2ac\cos{\beta}.$$

Proof. Let $D$, $E$ and $F$ be the contact points of the incircle with $AC$, $AB$ and $BC$, respectively. Also, let $AE=AD=x$; $BE=BF=y$; $CD=CF=z$. We start from two well-known relationships of a triangle: $$\sin^2{\frac{\gamma}{2}}=\frac{(s-a)(s-b)}{ab} \qquad\text{and}\qquad \cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}$$  
(See Cut-the-knot's Relations between various elements of a triangle for proofs), where $s$ denotes the semiperimeter of $\triangle{ABC}$. Since $(s-a)=x$, $(s-b)=y$ and $(s-c)=z$, then the following identity holds:
$$ab\cos{\gamma}=ab\cos^2{\frac{\gamma}{2}}-ab\sin^2{\frac{\gamma}{2}}=sz-xy$$
Substituting and multiplying by 4, 
$$4ab\cos{\gamma}=(a+b+c)(a+b-c)-(b+c-a)(a+c-b)$$
Simplifying,
$$2ab\cos{\gamma}=a^2+b^2-c^2$$
$\square$

A similar reasoning must show that $a^2=b^2+c^2-2bc\cos{\alpha}$ and $b^2=a^2+c^2-2ac\cos{\beta}$.

AcknowledgementMy sincerest thanks to Angina Seng for giving helpful comments which allowed me to simplify the proof.

Related material.

miércoles, 20 de mayo de 2020

Another Simple Proof of Johnson's Theorem

Other proofs can be found in cut-the-knot.org. See also Johnson's Three Circles Theorem Revisited and Johnson's theorem proof.

Johnson's theorem: Let three equal circles with centers $J_a$, $J_b$, and $J_c$ intersect in a single point $H$ and intersect pairwise in the points $A$, $B$, and $C$. Then the circumcircle of the triangle $\triangle{ABC}$ is congruent to the original three.


Proof. Since $\odot{ABH}$ and $\odot{BCH}$ are congruent and as $\angle{BAH}$ and $\angle{BCH}$ are angles subtended by the same arc, it follows that $\angle{BAH}=\angle{BCH}$. Analogously, $\angle{CAH}=\angle{CBH}$. Let $O$, $J_a$ be the centers of $\odot{ABC}$ and $\odot{BCH}$, respectively, then

$$\angle{COB}=2\angle{CAB}=2\angle{CAH}+2\angle{BAH}=2\angle{CBH}+2\angle{BCH}=\angle{BJ_aC}.$$

It follows that $\triangle{BCO}$ and $\triangle{BCJ_a}$ are similar isosceles triangles sharing a common side, $BC$, so by $ASA$ we deduce that $\triangle{BCO}\cong{\triangle{BCJ_a}}$. Consequently, $\odot{ABC}$ is congruent to the original three. 

$\square$

Note: The point $H$ may cross the side lines of the triangle $\triangle{ABC}$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.