Giải thích các bước giải:

Câu 15:

a.Vì $DC\perp AB\to \widehat{CIA}=\widehat{CIB}=90^o$

Mà $IA=IB\to\Delta CIA=\Delta CIB(c.g.c)$

b.Tương tự ta có : $\Delta DIA=\Delta DIB(c.g.c)\to\widehat{ADI}=\widehat{IDB}$

$\to DI$ là phân giác $\widehat{ADB}$

c.Từ câu a,b $\to CA=CB,DA=DB$

$\to\widehat{MAB}=\widehat{NBA}, \widehat{DAB}=\widehat{DBA}$

$\to\widehat{NAC}=\widehat{DAB}-\widehat{CAB}=\widehat{DBA}-\widehat{CBA}=\widehat{DBC}$

Lại  có $\widehat{NCA}=\widehat{MCB},CA=CB\to\Delta CNA=\Delta CMB(g.c.g)$

$\to CN=CM\to\widehat{CMN}=90^o-\dfrac 12\widehat{NCM}$

Mà  $CA=CB\to\widehat{CAB}=90^o-\dfrac 12\widehat{ACB}$

Vì $\widehat{NCM}=\widehat{ACB}\to \widehat{CMN}=\widehat{CAB}\to MN//AB$

Câu 16:

Ta có :
$\dfrac a5=\dfrac b3=\dfrac c2=k$

$\to a=5k, b=3k, c=2k\to 5k.3k=(2k)^2+11$

$\to 15k^2=4k^2+11$

$\to 11k^2=11$

$\to k^2=1$

$\to k=\pm 1$

$+)k=1\to a=5,b=3,c=2$

$+)k=-1\to a=-5,b=-3,c=-2$