10,20,30,40,60

10,20,30,40,60

Mesajde anamariaradu » Joi Ian 31, 2013 4:07 pm

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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Re: 10,20,30,40,60

Mesajde Vintu Vladimir » Mie Apr 17, 2013 2:16 pm

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
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Re: 10,20,30,40,60

Mesajde anamariaradu » Mie Apr 17, 2013 2:37 pm

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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Re: 10,20,30,40,60

Mesajde anamariaradu » Joi Apr 18, 2013 8:39 pm

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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Re: 10,20,30,40,60

Mesajde sunken rock » Vin Apr 19, 2013 9:27 am

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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Re: 10,20,30,40,60

Mesajde sunken rock » Vin Apr 19, 2013 10:32 am

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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Re: 10,20,30,40,60

Mesajde sunken rock » Mar Mai 30, 2017 6:28 pm

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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