In triunghiul \( ABC \) cu \( m(\widehat{ABC})<90\textdegree \), perpendiculara din \( A \) pe bisectoarea unghiului \( ABC \) intersecteaza dreapta \( BC \) in \( D \). Daca \( M \) este intersectia dintre perpendiculara in \( A \) pe \( AD \) si perpendiculara in \( B \) pe \( BD \), iar \( N \) intersectia dintre perpendiculara in \( B \) pe \( AB \) si perpendiculara in \( D \) pe \( AD \), sa se demonstreze ca \( MA=ND \) si \( MD=AN \).
Probleme date la olimpiade, RMT 1/1998
O.VI.30
Moderators: Bogdan Posa, Laurian Filip
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Andi Brojbeanu
- Bernoulli
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in \( \triangle ABD \) AD bisectoare si inaltime \( \Longrightarrow \triangle ABD \) isoscel, \( AB=BD \), \( \angle BAD\equiv \angle BDA(1) \).
\( m(\angle MAD)=90-m(\angle BAD)(2) \)
\( m(\angle BDN)=90-m(\angle ADB)(3) \)
-din \( (1),(2),(3) \Longrightarrow \angle MAD\equiv \angle BDN \).
\( m(\angle MBA)=90-m(\angle ABC)(4) \)
\( m(\angle NBD)=90-m(\angle ABC)(5) \)
-din \( (4),(5) \Longrightarrow \angle MBA\equiv\angle NBD \)
\( \triangle MAB\equiv \triangle NDB(LUL)\Longrightarrow MA=DN,BM=BN \)
\( \triangle MBD\equiv \triangle ABN(C.C)\Longrightarrow MD=AN \).
\( m(\angle MAD)=90-m(\angle BAD)(2) \)
\( m(\angle BDN)=90-m(\angle ADB)(3) \)
-din \( (1),(2),(3) \Longrightarrow \angle MAD\equiv \angle BDN \).
\( m(\angle MBA)=90-m(\angle ABC)(4) \)
\( m(\angle NBD)=90-m(\angle ABC)(5) \)
-din \( (4),(5) \Longrightarrow \angle MBA\equiv\angle NBD \)
\( \triangle MAB\equiv \triangle NDB(LUL)\Longrightarrow MA=DN,BM=BN \)
\( \triangle MBD\equiv \triangle ABN(C.C)\Longrightarrow MD=AN \).