ta co $\widehat{DAB}=\widehat{BED}=\widehat{BFD}=90^0$$=> $ ngu giac ABEDF noi tiep=>$\widehat{BFA}=\widehat{BEA}$
ma $\widehat{NAF}=(\widehat{NAC}+\widehat{CAF})_1=(\widehat{ACE}+\widehat{CAE})_2=\widehat{AEB}=\widehat{AFB}$
$_1= _2$ vi tam giac AMC can tai M=>$\widehat{NAC}=\widehat{ACE}$
$\widehat{CAF}=\widehat{CAE}$ vi chieu hai cung DE=DF
tu cac dieu tren =>tam giac NAF can tai N +>NA=NF