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Posted: **Sun Dec 14, 2008 5:59 pm**

\( \triangle\ ABC\ \ :\ \ \left\|\ \begin{array}{ccc} D\in (BC) & , & \widehat {DAB}\equiv\widehat {DAC} \\
\ K\in AD & , & BK\perp AD \\
\ H\in BC & , & AH\perp BC \\
\ X\in AB\cap HK & , & Y\in AC\cap XD\end{array}\ \right\|\ \Longrightarrow\ \overline {\underline {\left\|\ YA = YC\ \Longleftrightarrow\ A = 90^{\circ}\ \right\|}} \)

Posted: **Sun Dec 14, 2008 7:54 pm**

Folosim :

Lema Daca AD este inaltime in triunghiul ascutitunghic ABC atunci \( \frac{BD}{DC}=\frac{\tan C}{\tan B} \)

Asadar daca presupunem ca \( A=90^{\circ} \) din teorema lui Menelaus \( \frac{AX}{XB}\cdot\frac{\tan D}{\tan B}\cdot\frac{\tan A}{\tan D}=1 \)

deci \( \frac{AX}{XB}=\tan B \) (1)

Tot din teorema lui Menelaus \( \frac{AX}{XB}\cdot\frac{AB}{AC}\cdot\frac{CY}{YA}=1 \) (2)

Din (1) & (2) rezulta ca CY=YA.

Reciproc analog presupunand ca CY=YA, obtinem ca \( \tan A=1 \)

mateforum.ro

YA=YC <=> A=90

YA=YC <=> A=90