m∡(MB, (ABC))=m∡(MB, AB)=45°,ptca [MA]≡[AB], ΔMAB dreptunghic isoscel
m∡(MD, (ABC))=m∡(MD,AD)=∡(MDA)
cum AD=BC=8√3 si MA=8, tg∡(MDA)=8/8√3=1/√3 deci ∡(MDA)=30°
m∡(MO, (ABC))=m∡(MOA)
AO=BD/2=(1/2) *√(8²+(8√3)²)=(1/2) *16=8
atunci si ΔMAO dreptunghic isioscel, m∡(MOA)=45°
as simple as that!
