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I thought about an "Object Oriented Geometry" (Geometria Orientata pe Obiecte) inspired by a programming language. I found among my old papers a sketch of such a project. It has NOT materialized, and will remain at this stage.
I present an EXAMPLE of such an Object here; I needed this configuration in Solving Problem E: 17499 .
Properties of this Object :
- $AB=BC=CD=DA=AE=AF \tag{1}$
- $\widehat{BAD}=\widehat{ABC}=\widehat{BCD}=\widehat{ADC}=90^{\circ} \tag{2}$
- $\widehat{BAE}=\widehat{ABE}=\widehat{AEB}=60^{\circ} \tag{3}$
< Object figure >
Starting from this Object $\textbf{O}$ , we will add new elements, obtaining other Objects. For example, by joining points $C\; and \;E$ , a new Object $\textbf{O}_{\textbf{1}}$ is obtained.
We will say that $\textbf{O}\; is\; a\; SubObject\; of\;\textbf{O}_{\textbf{1}}$. We could write, pedantically, like this
$\textbf{O}\subset \textbf{O}_{\textbf{1}}$.
The properties of $O_1$ are :
- Properties (1) - (3) of $O$ ;
- $\widehat{BEC}=\widehat{BCE}=15^{\circ} \tag{4}$
- $\widehat{AEC}=45^{\circ}\;,\;\;\widehat{AGE}=\widehat{DCE}=75^{\circ} \tag{5}$
Next, we obtain an object $\textbf{O}_{\textbf{2}}$ , joining points $D\; and\; B$. We denote by $ F$ a new intersection point.
< $\textbf{O}_{\textbf{2}}$ >
We no longer list the Properties of this new Object.
From Object $O_1$ a new object is obtained, whose angles we have put in the image.
< Angles in $O_3$ >
We notice that $O_3$ has two objects of type $O_1$ in its component.
Let us now consider the following figure
Indeed, triangles $ABF\; and \;CBF$ are congruent, being symmetrical with respect to the line $BD$ (or with the case SAS : $BA=BC\;,\;BF=BF\;,\;\widehat{ABF}=45^{\circ}=\widehat{CBF}$. Then, in triangle $BEF\;:\;\widehat{BFE}=180^{\circ}-\widehat{BEF}-\widehat{EBF}=$
$=180^{\circ}-15^{\circ}-(60^{\circ}+45^{\circ})=60^{\circ}$ ; in $AEF\;:\;\widehat{AFE}=180^{\circ}-\widehat{AEF}-\widehat{EAF}=180^{\circ}-(15^{\circ}+30^{\circ})-(60^{\circ}+15^{\circ})=$ $=60^{\circ}\;$; and around point $F\;:\;\widehat{AFD}=\widehat{BFD}-\widehat{AFE}-\widehat{BFE}=180^{\circ}-60^{\circ}-60^{\circ}=60^{\circ}$
Let us note that, since $\widehat{EAD}=60^{\circ}+90^{\circ}=150^{\circ}$ and$\widehat{EAF}=60^{\circ}+15^{\circ}=75^{\circ}=\frac{150^{\circ}}{2}\;$
then the line $AF$ , being the bisector of the angle $\measuredangle DAE$ at the vertex of the isosceles triangle $ADE$ , is the perpendicular bisector of the segment $[DE]$.
We will use these last considerations in solving Problem E:17499.
