Slicing A=100 AB=AC

Slicing A=100 AB=AC

Mesajde Marius Stănean » Lun Noi 25, 2013 7:24 pm

Fie \triangle ABC,\;AB=AC,\;\angle A=100^\circ. Pe prelungirea lui AB se ia punctul D astfel incat AD=BC. Sa se afle \angle ADC.

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Re: Slicing A=100 AB=AC

Mesajde anamariaradu » Dum Dec 01, 2013 3:49 pm

Solutia 1 :

Fie E un punct pe mediatoarea segmentului AD astfel incat triunghiurile AED si BAC sa fie congruente.(E si C in acelasi semiplan determinat de dreapta AD)
Atunci AE=AB=AC=DE si m(\angle EAC)=100-40=60 de unde triunghiul EAC este echilateral .
\implies DE=EC
In triunghiul isoscel DEC avem m(\angle DEC)= 100+60=160 de unde m(\angle CDE)= 10 si m(\angle ADC)=40-10=30^\circ.

Solutia 2 :

Fie E un punct in plan astfel incat triunghiul ADE este echilateral. (E si C in acelasi semiplan determinat de dreapta AD)
Avem : AD=AE=DE=BC si m(\angle EAC)=100-60=40 .
De aici triunghiurile DAC si ACE sunt congruente.
\implies m(\angle AEC)=40 , m(\angle DEC)=100 de unde triunghiurile DAC si DEC sunt congruente
\implies m(\angle ADC)=30^\circ
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Re: Slicing A=100 AB=AC

Mesajde sunken rock » Dum Dec 01, 2013 5:12 pm

Amazing solutions!

Another one:
Extend (BA to M so that CM=AC, take N on (BC) so that CN=AC, see that \triangle CMN is equilateral, hence MN=CN, also \angle BMN=20^\circ\ (\ 1\ ) and BN=BD\ (\ 2\ ). From (2), with \angle ABC=40^\circ we infer \angle BDN=20^\circ. With (1), DN=MN, consequently N is the circumcenter of \triangle CMD, therefore \angle MDC=\frac{\widehat{MNC}}{2}=30^\circ.

Yet another one, shorter:
Construct the equilateral \triangle CBE, A and E on either side of BC. Then ABEC is a kite and, with \angle ACE=\angle BAC=100^\circ and CE=BC=AD, we get ADEC an isosceles trapezoid. By angle chase we infer DE=BE. We also see that \triangle DCA\cong\triangle EAC, with \angle ADC=\angle AEC=30^\circ.

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Re: Slicing A=100 AB=AC

Mesajde sunken rock » Mie Iun 13, 2018 9:11 pm

Another nice idea:
Construct the equilateral triangle $\triangle ACO$, $O,B$ on the same side of $AC$, then $\triangle OAD\equiv\triangle ABC$ (s.a.s.), so $\angle AOD=100^\circ$ and $OD=OC$, therefore $O$ is the circumcenter of $\triangle ADC$, wherefrom $\angle ODC=10^\circ$ and consequently $\angle ADC=30^\circ$.

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