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TMMATE problema 4

Posted: Mon Jan 26, 2009 3:29 pm
by Beniamin Bogosel
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) \( \frac{AB}{AC}+\frac{AC}{AB}=4 \).

TMMATE 2009

Posted: Mon Feb 02, 2009 12:16 am
by Marius Mainea
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).