10,20,30,40,60

anamariaradu
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Membru din: Lun Aug 06, 2012 3:35 pm

10,20,30,40,60

Mesaj de anamariaradu »

Fie $ABC$ un triunghi cu $E ,D$ puncte pe $AB$ respectiv $AC$. Daca masurile unghiurilor $BAC$, $ACE, BCE,$ si $ABD$ sunt $20^\circ, 30^\circ, 10^\circ,$ respectiv $60^\circ$ gasiti $m(\angle BDE)$.
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Vintu Vladimir
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Re: 10,20,30,40,60

Mesaj de Vintu Vladimir »

vom demonstra ca $\angle{BDE}=20^\circ$
pentru inceput, din teorema sinusurilor in $\triangle{BEC}\Rightarrow\dfrac{BE}{\sin{10^\circ}}=\dfrac{a}{\cos{40^\circ}}$, deci $BE=\dfrac{a\sin{10^\circ}}{\cos{40^\circ}}$
cum $\dfrac{AD}{\sin{60^\circ}}=\dfrac{c}{\sin{80^\circ}}\Rightarrow AD+BE=\dfrac{c\sin{60^\circ}}{\sin{80^\circ}}+\dfrac{a\sin{10^\circ}}{\cos{40^\circ}}$
vom demonstra acum ca $AD+BE=AB\Leftrightarrow\dfrac{\sin{60^\circ}}{\sin{80^\circ}}+\dfrac{a}{c}\cdot\dfrac{\sin{10^\circ}}{\cos{40^\circ}}=1$ $\Leftrightarrow-2\sin{10^\circ}\sin{20^\circ}=\sin{60^\circ}-\sin{80^\circ},(A)$ din $-2\sin{10^\circ}\sin{20^\circ}=\cos{10^\circ}-\cos{30^\circ}$
astfel $\Rightarrow AD+BE=AB$, adica $AD=AE\Rightarrow\angle{ADE}=80^\circ$, de unde obtinem $\angle{BDE}=20^\circ$
anamariaradu
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Membru din: Lun Aug 06, 2012 3:35 pm

Re: 10,20,30,40,60

Mesaj de anamariaradu »

Altfel:
Prelungim $AB$ din $B$ pana in $F$ astfel incat $< EFD = 30^\circ$ .Obtinem $< CEF = 50^\circ$, si $< CDF = 50^\circ ,DF\perp BC.$
Avem $< BDF = 30^\circ$, si in final in patrulaterul inscriptibil CDEF :$x + 30^\circ = 50^\circ$ de unde $x = 20^\circ.$
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anamariaradu
Mesaje: 251
Membru din: Lun Aug 06, 2012 3:35 pm

Re: 10,20,30,40,60

Mesaj de anamariaradu »

Altfel:
Fie $F$ simetricul lui $D$ fata de $CE.$ Triunghiul $DFC$ este echilateral.
In patrulaterul $DCFB$ avem $m(\angle DBC)=m(\angle DFC)=60$ ,deci patrulaterul $DCFB$ este inscriptibil.
$m(\angle ABD)=m(\angle FCD)=60,$ deci $A,E,B, F$ sunt coliniare.
Avem $m(\angle AFD)=m(\angle BCD)=40.$ Dar triunghiul $FED$ este isoscel , deci $m(\angle EDF)=40.$
In final $m(\angle EDB)=40-20=20.$
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sunken rock
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Re: 10,20,30,40,60

Mesaj de sunken rock »

Outline of another solution: Reflect $B$ across $AC$ to $P$ and take $Q\in AB\cap PD. PCBQ$ is cyclic and $\widehat{BPD}=10^\circ$... $Q\equiv E$.

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sunken rock
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sunken rock
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Re: 10,20,30,40,60

Mesaj de sunken rock »

Yet another one:
Take $O$, the circumcenter of $\Delta BCE$; it belongs to $AC$ and $BODE$ is an isosceles trapezoid (angle chase), a.s.o.

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sunken rock
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sunken rock
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Re: 10,20,30,40,60

Mesaj de sunken rock »

Let the perpendicular bisector of $AC$ be cut by $CE,CB$ at $F,G$ respectively. Then $AFBC$ is cyclic and $\measuredangle EFB=20^\circ\ (\ 1\ )$. Since the reflection of $D$ across $AB$ belongs both to $AG$ and $BC$, we conclude that $G$ is that reflection. But $GFEB$ is cyclic from $\angle GBE+\angle GFE=60^\circ+120^\circ=180^\circ$, wherefrom $\angle BDE=\angle BGE=\angle BFE=20^\circ$.

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sunken rock
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