Conc.interj."Grigore Moisil" Urziceni 2010 probl.1

[Andi Brojbeanu]

In tetraedrul \( ABCD \) trideptunghic in \( D \), unghiul \( \angle{BAD} \) su unghiul \( \angle{ACD} \) au aceasi masura, iar \( H \) este punctul de intersectie al inaltimilor \( CE \) si \( BF \) ale triunghiului \( \bigtriangleup{ABC} \).
Aratati ca: \( EH\cdot EC+ FH\cdot FB= BD\cdot DC \).
Profesor Paunescu Constantin, Urziceni

[Andi Brojbeanu]

\( ABCD \) tetraedru tridreptunghic \( \Rightarrow DC \perp DA, DC\perp DB\Rightarrow DC\perp (DAB)\Rightarrow DC\perp AB\Rightarrow AB\perp DC \).
Din \( AB\perp DC \) si \( AB\perp CE\Rightarrow AV\perp (DEC)\Rightarrow AB\perp DH\Rightarrow DH\perp AB \). Analog, \( DH\perp AC\Rightarrow DH\perp (ABC) \)
\( \Rightarrow DH \) inaltime in triunghiurile dreptunghice \( DEC \) si \( DBF \)\( \stackrel{\small T.Catetei}{\Rightarrow}EH\cdot EC=DE^2 \) si \( FH\cdot FB=DF^2 \).
Triunghiurile dreptunghice \( DAB \) si \( DCA \) sunt asemenea (conform ipotezei \( m(\angle{BAD})=m(\angle{ACD}) \)), deci: \( \frac{AD}{DC}=\frac{DB}{DA}\Rightarrow AD^2=BD\cdot DC \).
Asadar, trebuie sa demonstram ca: \( DE^2+DF^2=AD^2 \).
Cu notatiile \( AD=a, DB=b, DC=c \) si aplicand teorema inaltimii in triunghiurile dreptunghice \( ADB \) si \( ADC \) relatia devine:
\( (\frac{ab}{\sqrt{a^2+b^2}})^2+(\frac{ac}{\sqrt{a^2+c^2}})^2=a^2\Leftrightarrow \frac{a^2b^2}{a^2+b^2}+\frac{a^2c^2}{a^2+c^2}=a^2|:a^2\Leftrightarrow \frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}=1\Leftrightarrow \frac{b^2}{a^2+b^2}=\frac{a^2+c^2-c^2}{a^2+c^2}\Leftrightarrow \frac{b^2}{a^2+b^2}=\frac{a^2}{a^2+c^2} \)
\( \Leftrightarrow a^2b^2+b^2c^2=a^4+b^2a^2\Leftrightarrow a^4=b^2c^2\Leftrightarrow a^2=bc\Leftrightarrow AD^2=BD\cdot DC \), adevarat.