a) MA _|_ (ABC) => MA _|_ BC (1).
AB_|_BC (pentru ca ABCD-dreptunghi) (2).
Din (1) si (2) => BC _|_ (MAB) => MB _|_ BC (3).
Din (2) si (3) => m<((MBC),(ABC))=m(<MBA)=30*.
tg(<MBA)=[tex] \frac{MA}{AB}= \frac{ \sqrt{3} }{3} [/tex] => [tex]AB= \frac{3*MA}{ \sqrt{3} }= \frac{36}{ \sqrt{3} }= \frac{36 \sqrt{3} }{3}= 12\sqrt{3}(cm). [/tex]
CD_|_AD si CD_|_AM => CD _|_ (AMD) => CD _|_ MD (4).
AD_|_CD (5).
Din (4) si (5) => m<((MDC),(ABC)=m(<MDA)=60*.
tg(<MDA)=[tex] \frac{MA}{AD}= \sqrt{3} =>AD= \frac{MA}{ \sqrt{3} }= \frac{12}{ \sqrt{3} }= \frac{12 \sqrt{3} }{3}=4 \sqrt{3} (cm). [/tex]
AD=BC=[tex]4 \sqrt{3} [/tex]cm.
b) Fie E∈BD a.i. AE _|_ BD.
MA_|_(ABC) |
MA_|_AE | => ME _|_ BD (R.2.T.3_|_)
AE_|_BD |
AE,DB⊂(ABC) |
d(M,BD)=ME
Din T. Pitagora in ΔABD se obtine BD=[tex]4 \sqrt{30} [/tex].
[tex]AE= \frac{AD*AB}{BD}= \frac{ 6\sqrt{10} }{5} (cm).[/tex]
Se aplica T. Pitagora in ΔMAE si se obtine ME=[tex]6 \sqrt{22} [/tex](cm).
Se duce inaltimea AX in triunghiul ΔAEM.
BD _|_ AE si BD _|_ ME => BD _|_ (AEM) => BD _|_AX, dar AX _|_ ME => AX _|_ (MBD).
AX se afla din triunghiul AEM:
[tex]AX= \frac{AM*AE}{ME} [/tex]=(calcule)
b) unghiul este <MEA
tg(<MEA)=[tex] \frac{AM}{AE}= \frac{12}{ \frac{6 \sqrt{10} }{5} }= \sqrt{10} . [/tex]
