Concurenta intr-un triunghi.

[Virgil Nicula]

Consideram \( \triangle\ ABC \) si \( D\in (BC) \) . Notam \( \left\|\ \begin{array}{ccc}
E\in AC & ; & DE\perp AC\\\\
F\in AB & ; & DF\perp AB\end{array}\ \right\| \)

si \( L\in BE\ \cap\ CF \) . Sa se arate ca \( AL\perp BC \ \Longleftrightarrow\ \widehat {DAB}\equiv\widehat {DAC} \) .

[Marius Mainea]

Din teorema lui Menelaus \( \frac{BM}{MC}=\frac{AE}{EC}\cdot\frac{BF}{FA}=\frac{AD\cos\angle{DAC}}{DC\cos\angle{C}}\cdot\frac{BD\cos\angle{B}}{AD\cos\angle{DAB}}=\frac{BD}{DC}\frac{\cos\angle{DAC}}{\cos\angle{DAB}}\frac{\cos B}{\cos C} \)

Asadar \( AL\perp BC \Longleftrightarrow\frac{BM}{MC}=\frac{BD\cos B}{DC\cos C}\Longleftrightarrow\cos\angle{DAB}=\cos\angle{DAC}\Longleftrightarrow\angle{DAB}=\angle{DAC} \)