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Problema de slicing (90,50,10)
Scris: Sâm Apr 30, 2011 8:57 am
de 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 $\angle CED\,.$
Re: Problema de slicing (90,50,10)
Scris: Sâm Apr 30, 2011 12:11 pm
de 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 $\widehat{CAE}=40$.
Re: Problema de slicing (90,50,10)
Scris: Dum Mai 01, 2011 2:19 pm
de 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
Re: Problema de slicing (90,50,10)
Scris: Lun Mai 22, 2017 5:36 pm
de 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