In triunghiul ABC in care \( AB<AC \), \( [AD \) este bisectoarea unghiului \( A(D\in BC) \). Perpendciulara in \( A \) in \( AD \) intersecteaza dreapta \( BC \) in \( M \). Pe dreapta \( AM \) consideram punctul \( N \) astfel incat \( AM=AN \).
a) Aratati ca \( \triangle{AMD}\equiv\triangle{AND} \).
b) Daca \( AC\cap ND={\{E\}} \) este adevarat ca \( AB=AE \)?
c) Demonstrati ca \( AD\perp BE \).
Probleme date la olimpiade, RMT 1/1998
O.VI.28
Moderators: Bogdan Posa, Laurian Filip
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Andi Brojbeanu
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O.VI.28
Last edited by Andi Brojbeanu on Mon May 11, 2009 1:59 pm, edited 1 time in total.
\( AM=AN \)
\( AD=AD \)
\( m(\angle MAD)=m(\angle DAN)=90 \)
- din relatiile de mai sus\( \Longrightarrow \triangle MAD\equiv \triangle NAD(C.C.) \)
- in \( \triangle MDN \), DA inaltime si mediana\( \Longrightarrow \triangle \) este isoscel, \( \angle BMA\equiv\angle ENA \).
\( \angle MAB=90-\angle BAD=90-\angle DAC=\angle EAN \)
\( \triangle MAB\equiv\triangle NAE(U.L.U)\Longrightarrow AB=AE, \triangle ABE \) isoscel.
- in \( \triangle ABE \) isoscel, AD bisectoare \( \Longrightarrow \) AD inaltime, \( AD\perp BE \).
\( AD=AD \)
\( m(\angle MAD)=m(\angle DAN)=90 \)
- din relatiile de mai sus\( \Longrightarrow \triangle MAD\equiv \triangle NAD(C.C.) \)
- in \( \triangle MDN \), DA inaltime si mediana\( \Longrightarrow \triangle \) este isoscel, \( \angle BMA\equiv\angle ENA \).
\( \angle MAB=90-\angle BAD=90-\angle DAC=\angle EAN \)
\( \triangle MAB\equiv\triangle NAE(U.L.U)\Longrightarrow AB=AE, \triangle ABE \) isoscel.
- in \( \triangle ABE \) isoscel, AD bisectoare \( \Longrightarrow \) AD inaltime, \( AD\perp BE \).