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Concursul Matefbc editia a 3-a problema 4
Posted: Sun Dec 06, 2009 9:08 pm
by Andi Brojbeanu
Fie \( ABCD \) un patrulater inscriptibil, iar \( M \) un punct pe diagonala \( AC \). Cercul circumscris triunghiului \( BCM \) intersecteaza dreapta \( AB \) in \( P \), iar cercul circumscris triunghiului \( MCD \) taie dreapta \( AD \) in \( L \). Sa se arate ca:
a) Punctele \( P, M, L \) sunt coliniare.
b) Punctele \( B, P, L, D \) sunt conciclice.
Posted: Sun May 23, 2010 2:55 pm
by Andi Brojbeanu
a) Din inscriptibilitatea parulaterelor \( ABCD \),\( BCMP \) si \( DCML \) rezulta ca:
\( m(\angle{ABC})=180\textdegree-m(\angle{ADC}) \) si \( m(\angle{BCA})=m(\angle{BDA})(1) \)
\( m(\angle{PMC})=180\textdegree-m(\angle{PBC})(2) \) si \( m(\angle{LMC})=180\textdegree-m(\angle{LDC}) \).
Asadar, \( m(\angle{PML})=m(\angle{PMC})+m(\angle{LMC})=360\textdegree-[m(\angle{PBC})+m(\angle{LDC})]=360\textdegree-180\textdegree=180\textdegree\Rightarrow \) punctele \( P, L, M \) sunt coliniare.
b) Din \( (1) \) si \( (2)\Rightarrow m(\angle{PBD})+m(\angle{BDL})=m(\angle{BPM})+m(\angle{BCM})=180\textdegree\Rightarrow \) patrulaterul \( BDLP \) inscriptibil \( \Rightarrow \) punctele \( B, P, L, D \) conciclice.