a) Xét \(\Delta\)AMN và \(\Delta\)CPNcó:
AN = NC )gt)
\(\widehat{ANM}=\widehat{PNC}\) (đối đỉnh)
MN = NP (gt)
=> \(\Delta\)AMN= \(\Delta\) CPN (c.g.c)
=> AM = CP hay BM = CP
b) Vì \(\Delta\)AMN= \(\Delta\) CPN
=> \(\widehat{MAN}=\widehat{NCP}\)
=> AM // CP
=> \(\widehat{BMC}=\widehat{MCP}\) (so le trong)
Xét \(\Delta\)BMC và \(\Delta\) PCM có:
BM = PC
\(\widehat{BMC}=\widehat{MCP}\)
CM:chung
=> \(\Delta BMC=\Delta PCM\left(c.g.c\right)\) (1)
c) từ b => MP = BC
=> 2MN= BC
hay \(MN=\dfrac{1}{2}BC\)
(1) => \(\widehat{MCB}=\widehat{PMC}\) => MN//BC