Đáp án: $B$
Giải thích các bước giải:
Ta có: $BM.BC=BA^2(=4)$
$\to \dfrac{BM}{BA}=\dfrac{BA}{BC}$
$\to \Delta BMA\sim\Delta BAC(c.g.c)$
$\to \widehat{BMA}=\widehat{BAC},\widehat{BAM}=\widehat{BCA}$
Mà $AM\perp MC$
$\to \widehat{NMC}=90^o-\widehat{AMB}=90^o-\widehat{BAC}=\widehat{ACB}$
$\to \widehat{NMC}=\widehat{NCM}$
$\to \Delta NMC$ cân tại $N$
$\to NM=NC=AC-AN=AC-kAC=(1-k)AC$ vì $\vec{AN}=k\vec{AC}$
Do $AM\perp MN$
$\to AM^2+MN^2=AN^2$
$\to AB^2+BM^2+MN^2=AN^2$
$\to 2^2+1^2+((1-k)AC)^2=(kAC)^2$
$\to 5+(1-k)^2AC^2=k^2AC^2$
Ta có $AC^2=AB^2+BC^2=20$
$\to 5+20(1-k)^2=20k^2$
$\to 20k^2-40k+25=20k^2$
$\to 40k=25$
$\to k=\dfrac58$
