Slicing A=100 AB=AC

Marius Stănean
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Slicing A=100 AB=AC

Mesaj de Marius Stănean »

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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anamariaradu
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Re: Slicing A=100 AB=AC

Mesaj de anamariaradu »

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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sunken rock
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Re: Slicing A=100 AB=AC

Mesaj de sunken rock »

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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sunken rock
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sunken rock
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Re: Slicing A=100 AB=AC

Mesaj de sunken rock »

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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sunken rock
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sunken rock
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Re: Slicing A=100 AB=AC

Mesaj de sunken rock »

... Or construct the isosceles triangle ADO with $\widehat{AOD}=100^\circ$, C and O on the same side of AB; as triangles ADO and ABC are congruent (s.a.s.) it follows that triangle AOC is equilateral wherefrom O is the circumcenter of triangle ADC and $2m(\widehat{ADC})=m(\widehat{AOC})=60^\circ$.

Best regards,
sunken rock
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