Problema de slicing (90,50,10)

[Marius Stănean]

In $\triangle ABC,\angle A=90^\circ,\angle B=50^\circ,D\in(BC),E\in(AD)$ astfel incat $\angle ACE=10^\circ, \,CE=AB$. Gasiti $\angle CED\,.$

[BocanuMarius]

Aratam ca $\widehat{CAE}=40$, de unde $\widehat{CED}=50$. Notam $BA=x$ si exprimam $AC,CE,CP,PE,AP$ doar in functie de $x$ si tangente de unghiuri.$AC=xtg50$,$CE=x$,$AP=xtg50tg10$,$CP=\frac{xtg50}{cos10}$, $[tex]$PE=x(\frac{tg50}{cos10}-1[\tex]. $\frac{CE}{PE}=$$\frac{AP}{AC}\frac{sin\widehat{PAE}}{sin\widehat{EAP}}$ si aratam doar prin calcule simple trigonometrice ca $\widehat{CAE}=40$.

[sunken rock]

Just take $B^\prime$ the reflection of $B$ in $A$, see that $AB^{\prime}CE$ is an isosceles trapezoid, hence $AE$ is median of $\triangle ABC$.

Best regards,
sunken rock

[sunken rock]

Remark: the problem can be generalized, as follows:
If $m(\widehat{ABC})=\alpha> 45^\circ$ and $m(\widehat{ACE})=2\alpha-90^\circ$, $CE=AB$, gives $\triangle ABD$ isosceles.

Best regards,
sunken rock