a) fie D∈BC, BD=DC, ⇒AD⊥BC
MA⊥(ABC), AD⊥B
AD, BC⊂(ABC)⇒(T3p)MD⊥BC⇔d(M, BC)=MB
MA=12
AD=24√3/2=12√3
MD=√MA²+AD²=√(12²+(12√3)²)=24
b) D, (A, (MBC))=AP, unde AP⊥MD, P∈MD
AP inalt coresp ipotenuzei in tr dr MAB
AP=AM*AB/MB=12*12√3/24√3=6√3
sau , altfelAP=AB sin30°=12√3/2=6√3 (pt ca mas B=30°, pt ca AM=MB/2)
c)
(MAD)∩(MAC)=MA
MA⊥(ABC)⇒MA⊥AD, AD⊥MA, AD⊂(MAD)
⇒MA⊥AC, AC⊥MA, AC⊂(MAC)
⇒m ∡((MAD)< (MAC))=m∡(AD, AC)=60°/2= 30° (in tr echilat inaltimea e si bisectoare)
