Fie AO inaltimea tetraedului
AO⊥(BCD)
∡((ADC),(BDC))=∡AFB=∡AFO
(ADC)∩(BDC)=DC
AO⊥BF,
BF⊂(BDC)
AF⊥DC
AF⊂(ADC)
OF⊥DC
muchia=2a
BF inaltime in ΔBDC echilateral⇒
[tex]BF=\frac{l\sqrt{3} }{2} =a\sqrt{3}[/tex]
OF se afla la o treime de baza
[tex]OF=\frac{1}{3} \times BF=\frac{a\sqrt{3} }{3}[/tex]
AF=BF ( ABCD tetraedru)
[tex]cosAFO=\frac{FO}{AF} =\frac{\frac{a\sqrt{3} }{3} }{a\sqrt{3} } =\frac{1}{3}[/tex]
AE=AF=a√3
Luam ΔAEF isoscel
Fie AM⊥EF, M mijloc
[tex]MF=\frac{a}{2}[/tex]
AM²+MF²=AF²
[tex]AM^2=3a^2-\frac{a^2}{4}=\frac{11a^2}{4}\\\\ AM=\frac{a\sqrt{11} }{2}[/tex]
Fie EN⊥AF
EN×AF=AM×EF
[tex]EN=\frac{a\sqrt{11} }{2} \times a\times \frac{1}{a\sqrt{3} } =\frac{a\sqrt{33} }{6}[/tex]
[tex]sinEAF=sinEAN=\frac{EN}{AE} =\frac{\frac{a\sqrt{33} }{6} }{a\sqrt{3} }=\frac{\sqrt{11} }{6}[/tex]
