Karede Bir Açı ve Bir ... Hiçbir Yerden Uçurtma

           An Angle in a Square and a Kite Out of Nowhere

          Uçurtma, akademik dilde Deltoid olarak da adlandırılır.

          Starting from a geometry problem and the solution that Hasan Ata offered, we formulate the following problem:

          " Let $ABCD$ be a square and an angle $\angle EAF$ with $E \in ]BC[,\;F \in ]CD[$. Prove that 

$$\measuredangle EAF=45^{\circ}\;\Leftrightarrow\;EF=BE+DF."$$

Solution CiP

                    Let $\measuredangle EAF=45^{\circ}$

               Rotate the triangle $ABE$ with  $90^{\circ}$ clockwise around point $A$ and obtain the triangle $ADG$. We have $\measuredangle EAG=90^{\circ}$, $AE=AG$ and $DG=BE$.

  Since $\measuredangle FAG=90^{\circ}-45^{\circ}=45^{\circ}=\measuredangle EAF$ , it follows that the line $AF$  is the angle bisector of the $\angle EAG$ in the isosceles triangle $AEG$ so it is the perpendicular bisector of the segment $[EG]$, hence $FG=FE$, so $EF=FG=DG+DE=BE+DF$.

q.e.d.

           Reciprocally, let $EF=BE+DF$

       The same rotation now shows us that $FG=FE$ so quadrilateral $AEFG$ is a kite.

  Hence line $FA$ is simultaneously angle bisector of $\angle EFG$ and $\angle EAG$. Because $\measuredangle EAG=90^{\circ}$ it follow $\measuredangle EAF=\frac{90^{\circ}}{2}=45^{\circ}.$ 

q.e.d.

$\blacksquare$