Se considera un triunghi \( ABC \) in care \( m(\angle ABC)=75^o \). Fie \( D \in (BC) \) astfel incat \( AD \perp BC \). Stiind ca \( AD=\frac{1}{4}BC \), aratati ca
a) triunghiul este dreptunghic
b) \( \frac{AB}{AC}+\frac{AC}{AB}=4 \).
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Fie \( E\in [AD] \) astfel incat \( \angle(EBD)=60^{\circ} \)
Notand \( BD=x \) rezulta \( AD=(2+\sqrt{3})x \) si \( DC=(7+4\sqrt{3})x \)
Asadar \( \triangle ABD\sim\triangle CAD \) de unde \( \angle(BAC)=90^{\circ} \)
Deasemenea tot din asemanare \( \frac{AB}{AC}=\frac{BD}{AD}=\frac{1}{2+\sqrt{3}} \) si de aici rezlta si punctul b).
Notand \( BD=x \) rezulta \( AD=(2+\sqrt{3})x \) si \( DC=(7+4\sqrt{3})x \)
Asadar \( \triangle ABD\sim\triangle CAD \) de unde \( \angle(BAC)=90^{\circ} \)
Deasemenea tot din asemanare \( \frac{AB}{AC}=\frac{BD}{AD}=\frac{1}{2+\sqrt{3}} \) si de aici rezlta si punctul b).