mateforum.ro

Se considera punctele necoplanare A,B,C,D astfel incat:

\( AB=BD=CD=AC=\sqrt{2}AD=\frac{\sqrt{2}}{2}BC=a. \)

Aratati ca:

a) Exista un punct pe segmentul [BC] egal departat de punctele A,B,C,D.
b)\( 2m(\angle{(AD,BC)})=3m(\angle{((ABC),(BCD))}). \)
c) \( 6[d(A,DC)]^2=7[d(A,BCD)]^2 \)

a) \( BC=a\sqrt{2} , \Rightarrow \Delta BDC \) si \( \Delta ABC \) dreptunghice . Fie \( M \) mijlocul lui \( BC \Rightarrow MD=\frac{a\sqrt{2}}{2} \) si \( AM=\frac{a\sqrt{2}}{2} \Rightarrow MA=MB=MC=MD=\frac{a\sqrt{2}}{2} \)

b) \( \angle((ABC)(BCD))=60 \)
\( BC\perp DM \) si \( BC\perp AM \Rightarrow BC \perp(ADM) \Rightarrow \angle (BC;AD)=90
\Rightarrow 2\cdot90=3\cdot 60 \)

c) Fie \( CT \perp AD \Rightarrow CT^2=\frac{14a^2}{16} \Rightarrow CT=\frac{a\sqrt{14}}{4} \)
Fie \( AH\perp CD \Rightarrow AH =\frac{a\sqrt{7}}{4} \)
Fie \( AQ\perp DM \Rightarrow AQ \perp (BCD) \Rightarrow AQ=\frac{a\sqrt{6}}{4} \)
\( \Rightarrow 6\cdot\frac{7a^2}{16}=7\cdot\frac{6a^2}{16} \)