A L I N A : Un Obiect Geometric

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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 .

< O >

The OBJECT consists of the square  $ABCD$  and the equilateral triangle  $ABE$.

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.

   < $\textbf{O}_{\textbf{1}}$>

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  $O$ ;

- $\widehat{BEC}=\widehat{BCE}=15^{\circ} \tag{4}$

- $\widehat{AEC}=45^{\circ}\;,\;\;\widehat{AGE}=\widehat{DCE}=75^{\circ} \tag{5}$

<  Angles in  $O_1$ >

          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

We have the following values ​​of some angles in this 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.