Slicing A=20 (AB=AC) (own) reload

[Marius Stănean]

Fie $\triangle ABC (AB=AC),\;\angle A=20^\circ$, $D$ centrul cercului circumscris iar $E$ piciorul bisectoarei din $B$ pe $AC$. Sa se determine $\angle AED$.

[Stefan Dominte]

Pp ca $\triangle{DEC}$ nu e isoscel. Fie $M$ pe $AC$ a.i. $\triangle {DCM}$ sa fie isoscel $\Rightarrow \widehat{DMA}=20$
Aplicand teorema sinusurilor in ADM:
$\dfrac{AM}{DM}=\dfrac{AM}{MC}=\dfrac{sin30}{sin10}$.
Dar aplicand teorema sinusurilor in ABC:
$\dfrac{AB}{BC}=\dfrac{sin80}{sin20}$ $\Rightarrow$ $\dfrac{AM}{MC}=\dfrac{AB}{BC}$ $\Rightarrow$ $BM-bis\widehat{ABC}$ $\Rightarrow$ $M=E$ $\Rightarrow \widehat{DEA}=20$

[Stefan Tudose]

Sau notand cu $x=\widehat{DEA}$ si aplicand teorema sinusurilor in triunghiul $\triangle{ABE}$, se obtine:

$\dfrac{\sin{x}}{\sin{(60^\circ+x)}}=2\sin{10^\circ}=\dfrac{\sin{20^\circ}}{\sin{80^\circ}}$$\Leftrightarrow \cos{(40^\circ+x)}=\cos{(80^\circ-x)}$$\Leftrightarrow x=20^\circ$

[sunken rock]

To avoid trig, we shall work backward:
Let $\triangle AFE$ isosceles with $\hat A=20^\circ$ and $D$ inside it such that triangle DEF be equilateral. Extend $(AF, (AE$ to $B$ and $C$ respectively such that $FB=EF=CE$. We see easily that $BD=AD$ and $\angle ABE=40^\circ$.
By the uniqueness of the construction we are done.

Best regards,
sunken rock