Три проблема с приблизително една и съща конфигурация

           Това са броеве 1334, 1363 и 1366, публикувани тук http://gogeometry.com/.

 

           Problem 1334 (reformulation)

            Let $E$ be a point on the side $]BC[$ of the square $ABCD$. $[AF],\;\;F \in ]CD[$ is the angle bisector of the $\angle DAE$. Prove that $AE=BE+DF$.

    
In relation to the original figure, with the changed notes

we have equal angles $\angle BCE = \angle DHE$ (from the parallelism of the $CB$ and $AH$ lines), and $\angle BCE=\angle EBG$, hence $\angle EBG = \angle GHB$ so $BGH$ is an isosceles triangle and then $BG=a$.

           Let's get back to our figure.

If we note $x^{\circ}=\measuredangle AEB$ and we build the triangle $ADG$ so that $\Delta ABE \equiv \Delta ADG$ (basically, we rotate the triangle $ABE$ with  $90^{\circ}$ around point $A$, counterclockwise), we have $\measuredangle AGD=x^{\circ}$ and $[AG] \equiv [AE]$.

     Now we calculate $\measuredangle BAE=90^{\circ}-x^{\circ}=\measuredangle GAD$ and from $\angle AEB \equiv \angle DAE$ (due to the parallelism $BE \parallel AD$) we have $\measuredangle DAF=x^{\circ}, \;\measuredangle DAF=\frac{x^{\circ}}{2}$. Now $\measuredangle GAF=\measuredangle GAD +\measuredangle DAF =(90^{\circ}-x^{\circ})+\frac{x^{\circ}}{2}=90^{\circ}-\frac{x^{\circ}}{2}$. In the triangle $AGF$: $\measuredangle AFG=180^{\circ}-\measuredangle AGF-\measuredangle GAF=180^{\circ}-x^{\circ}-(90^{\circ}-\frac{x^{\circ}}{2})=90^{\circ}-\frac{x^{\circ}}{2}$. 

     From $\angle AFG \equiv \angle GAF$ it turns out that triangle $AFG$ is isosceles, with $GA=GF=a+b$.

$\blacksquare$

 

         Problem 1363

                   Solution CiP

          From $FC \perp BC,\;FG\perp BE$ and fact that $BF$ is bisector of $\angle CBE$ it follow $FC=FG$ so the congruence $\Delta BCF \equiv \Delta BGF$ and hence $BG=BC$. But $AB=BC$ so triangle $ABG$ is isosceles.

      Let $x^{\circ}=\measuredangle CBF=\measuredangle FBG$, then $\underline{\measuredangle GFB}=\measuredangle CFB=\underline{90^{\circ}-x^{\circ}}$.

Now, $\measuredangle ABG=90^{\circ}-2x^{\circ}$, so $\measuredangle BAG=\underline{\measuredangle BGA}=\underline{45^{\circ}+x^{\circ}}$.

      We calculate now $\measuredangle FGH=180^{\circ}-\measuredangle FGA=180^{\circ}-\measuredangle FGB-\measuredangle BGA=$

$=180^{\circ}-90^{\circ}-(45^{\circ}+x^{\circ})=45^{\circ}+x^{\circ}$.

But $\measuredangle BFG=\measuredangle FHG +\measuredangle FGH$ (as the outer angle of the triangle FGH) so

$\measuredangle FHG=\measuredangle BFG-\measuredangle FGH=(90^{\circ}-x^{\circ})-(45^{\circ}-x^{\circ})=45^{\circ}$.

 $\blacksquare$

 

            Problem 1366

                Solution CiP

             Consider the square $ABCD$ and the point $E$ on $]AD[$.

Line $BE$ intersects the circle once again at point $E'$ and the bisector $BF$ of the angle $\angle CBE$ intersects the circle again at point $F'$. Let $BE \cap AF' ={J}$.

      We will prove that $[BA] \equiv [BJ]$, so $J=G$ and so $F'=H$. (The fact that $BG$ is equal to the side of the square results easily from the congruence of the triangles $\Delta BCF \equiv \Delta BGF$.)

        First notice that point $F'$ is the middle of the arc $\overset{\frown}{C_DE'}\;:\;\overset{\frown}{CF'} \equiv \overset{\frown}{F'E'}$.

      Now applying the calculation formulas of the angles inscribed in the circle, respectively the angle with the tip inside the circle, we obtain 

$\measuredangle BAJ=\measuredangle BAF'=\frac{\overset{\frown}{B_CF'}}{2}=\frac{\overset{\frown}{BC}+\overset{\frown}{CF'}}{2}=\frac{90^{\circ}+\overset{\frown}{F'E'}}{2}=\frac{\overset{\frown}{AB}+\overset{\frown}{E'F'}}{2}=\measuredangle BJA.$

 It turns out that the triangle $ABJ$ is isosceles, with $BA=BJ$.

$\blacksquare$

           Remark Problem 1363 is now an immediate consequence of this result: $\measuredangle AHB=\frac{\overset{\frown}{AB}}{2}=\frac{90^{\circ}}{2}=45^{\circ}.$