Explicație pas cu pas:
a)
[tex]AD = BC = 3 \sqrt{6} \: cm[/tex]
T.P. în ΔABD dreptunghic:
[tex]{BD}^{2} = {AD}^{2} + {AB}^{2} = {(3 \sqrt{6})}^{2} + {(6 \sqrt{3})}^{2} \\ = 54 + 108 = 162[/tex]
[tex]BD = \sqrt{162} = > BD = 9 \sqrt{2} \: cm[/tex]
b) BE = BD - DE
[tex]BE = 9 \sqrt{2} - 3 \sqrt{2} = 6 \sqrt{2} \: cm[/tex]
[tex]DE \times BD = 3 \sqrt{2} \times 9 \sqrt{2} = 54 = {AD}^{2} \\ [/tex]
=> AE ⊥ BD
[tex]AE = \frac{AD \times AB}{BD} = \frac{3 \sqrt{6} \times 6 \sqrt{3} }{9 \sqrt{2}} = 6 \: cm \\ [/tex]
[tex]AM = MB = \frac{AB}{2} = 3 \sqrt{3} \: cm \\ [/tex]
[tex] \tg(BAE) = \frac{BE}{AE} = \frac{6 \sqrt{2} }{6} = \sqrt{2} \\ [/tex]
[tex]\tg(CMB) = \frac{BC}{MB} = \frac{3 \sqrt{6}}{3 \sqrt{3} } = \sqrt{2} [/tex]
=> ∢BAE ≡ ∢CMB
=> AE || CM
q.e.d.
