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Trisectie a unui unghi
Posted: Sun Jul 05, 2009 11:45 pm
by Virgil Nicula
Fie triunghiul \( ABC \) dreptunghic in \( A \) si punctele \( \{X,Y,Z,T\}\subset BC \) (linie !) astfel incat
\( \left\|\begin{array}{cccccc} B\in (XY) & , & Y\in (BC) & , & BX = BY = AB \\
\\
C\in (ZT) & , & Z\in (BC) & , & CZ = CT = AC\end{array}\right\| \).
Sa se arate ca \( \widehat {XAZ}\equiv\widehat {ZAY}\equiv\widehat {YAT}. \)
Posted: Mon Jul 06, 2009 10:53 am
by red_dog
\( \widehat{XAB}=\frac{B}{2}, \ \widehat{BAY}=90-\frac{B}{2}, \ \widehat{ZAC}=90-\frac{C}{2}, \ \widehat{CAT}=\frac{C}{2} \)
\( \widehat{XAT}=\frac{B}{2}+90+\frac{C}{2}=135 \)
\( 90=\widehat{XAB}+\widehat{BAY}=\widehat{XAZ}+\widehat{ZAY} \)
\( 90=\widehat{ZAC}+\widehat{CAT}=\widehat{ZAY}+\widehat{YAT} \)
Rezulta \( \widehat{XAZ}=\widehat{YAT} \)
Unghiurile \( \widehat{XAB} \) si \( \widehat{YAC} \) au laturile perpendiculare, deci \( \widehat{YAC}=\frac{B}{2} \)
Analog, \( \widehat{BAZ}=\frac{C}{2} \)
Atunci \( \widehat{ZAY}=\widehat{BAC}-\frac{B+C}{2}=45 \)
Cum celelalte doua unghiuri sunt egale si suma lor este 135, rezulta ca toate cele trei unghiuri sunt egale cu 45.
Metoda 2
Posted: Mon Jul 06, 2009 11:13 am
by Theodor Munteanu
\( \Delta XAY,\Delta ZAT \) sunt dreptunghice (mediana jumatate din ipotenuza), deci AZ, AY ceviene izogonale.
\( \frac{{{\rm ZA}}}{{{\rm AT}}} = tg(ATZ) = tg\left( {\frac{{180 - ACT}}{2}} \right) = tg\frac{C}{2} = \sqrt {\frac{{(p - a)(p - b)}}{{p(p - c)}}} = \sqrt {\frac{{(b + c - a)(a + c - b)}}{{(a + b + c)(a + b - c)}}} ; \\
ZY = ZC - YC; YC = BC - BY = a - c; ZY = b + c - a; YT = ZT - ZY = b + a - c; \\
\frac{{b + c - a}}{{b + a - c}} = \sqrt {\frac{{(b + c - a)(a + c - b)}}{{(a + b + c)(a + b - c)}}} \Leftrightarrow \sqrt {\frac{{b + c - a}}{{b + a - c}}} = \sqrt {\frac{{a + c - b}}{{a + b + c}}} \\ \) ceea ce rezulta imediat din Teorema lui Pitagora.
Deci \( \frac{{ZY}}{{YT}} = \frac{{ZA}}{{AT}} \), de unde AY e bisectoarea unghiului ZAT.