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O.VI.30

 
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Sun Apr 19, 2009 6:23 pm    Post subject: O.VI.30 Reply with quote

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
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salazar
Thales


Joined: 06 Apr 2009
Posts: 100
Location: Alba Iulia

PostPosted: Sun Apr 26, 2009 7:50 am    Post subject: Reply with quote

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.
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