Giải thích các bước giải:
a.Xét $\Delta AMN,\Delta CKN$ có:
$NA=NC$ vì $N$ là trung điểm $AC$
$\widehat{ANM}=\widehat{CNK}$
$NM=NK$
$\to\Delta AMN=\Delta CKN(c.g.c)$
$\to AM=CK$
Mà $M$ là trung điểm $AB\to MA=MB\to BM=CK$
b.Từ câu a
$\to\widehat{NAM}=\widehat{NCK}\to AM//CK\to AB//CK$
c.Xét $\Delta MKC,\Delta CMB$ có:
Chung $CM$
$\widehat{MCK}=\widehat{BMC}$ vì $CK//AB$
$CK=BM$
$\to\Delta MKC=\Delta CBM(c.g.c)$
$\to MK=CB$
Mà $NM=NK\to MK=2MN\to 2MN=BC\to MN=\dfrac12BC$
Bài 5:
Ta có:
$b^2=ac\to\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{2015b}{2015c}=\dfrac{a+2015b}{b+2015c}$
$\to\dfrac{a}{b}\cdot \dfrac{b}{c}=(\dfrac{a+2015b}{b+2015c})^2$
$\to \dfrac{a}{c}=(\dfrac{a+2015b}{b+2015c})^2$
