\(\overrightarrow{KA}=-\overrightarrow{AK}=-\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=-\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)

\(=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)

\(\overrightarrow{KD}=\overrightarrow{AD}-\overrightarrow{AK}=\overrightarrow{AD}+\overrightarrow{KA}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)

\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)

a: \(\overrightarrow{MC}=\frac13\cdot\overrightarrow{MB}\)

=>\(MC=\frac13MB\) và C nằm giữa M và B

MC+CB=MB

=>\(CB=MB-MC=MB-\frac13MB=\frac23MB\)

Ta có: \(\overrightarrow{NA}+3\cdot\overrightarrow{NC}=\overrightarrow{0}\)

=>\(\overrightarrow{NA}=-3\cdot\overrightarrow{NC}\)

=>N nằm giữa A và C và NA=3NC

NA+NC=AC

=>AC=NC+3NC=4NC

=>\(CN=\frac14CA\)

=>\(NA=3\cdot\frac14\cdot AC=\frac34AC\)

\(\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{0}\)

=>\(\overrightarrow{PA}=-\overrightarrow{PB}\)

=>P nằm giữa A và B và PA=PB

=>P là trung điểm của AB

=>\(AP=PB=\frac{AB}{2}\)

\(\overrightarrow{MP}=\overrightarrow{MB}+\overrightarrow{BP}\)

\(=-\overrightarrow{BM}+\overrightarrow{BP}=-\frac32\cdot\overrightarrow{BC}+\frac12\cdot\overrightarrow{BA}\)

\(=-\frac32\cdot\left(\overrightarrow{BA}+\overrightarrow{AC}\right)+\frac12\cdot\overrightarrow{BA}=-\frac32\cdot\overrightarrow{BA}-\frac32\cdot\overrightarrow{AC}+\frac12\cdot\overrightarrow{BA}\)

\(=-\overrightarrow{BA}-\frac32\cdot\overrightarrow{AC}=\overrightarrow{AB}-\frac32\cdot\overrightarrow{AC}\) (2)

\(\overrightarrow{NP}=\overrightarrow{NA}+\overrightarrow{AP}\)

\(=-\frac34\cdot\overrightarrow{AC}+\frac12\cdot\overrightarrow{AB}=\frac12\cdot\left(\overrightarrow{AB}-\frac32\cdot\overrightarrow{AC}\right)\) (1)

b: Từ (1),(2) suy ra \(\overrightarrow{NP}=\frac12\cdot\overrightarrow{MP}\)

=>N,M,P thẳng hàng

Xét ΔBAD có BI là đường trung tuyến

nên \(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)

=>\(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}\right)\)

\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{5}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{1}{3}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)=\dfrac{1}{6}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)=\dfrac{5}{6}\left(\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\right)\)

\(\overrightarrow{BM}=\overrightarrow{BA}+\overrightarrow{AM}\)

\(=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\)

=>\(\overrightarrow{BI}=\dfrac{5}{6}\cdot\overrightarrow{BM}\)

=>B,I,M thẳng hàng

Cách 1: Dùng định lý Menelaus đảo:

Từ đề bài, ta có \(\dfrac{BD}{BC}=\dfrac{2}{3}\), \(\dfrac{MC}{MA}=\dfrac{3}{2}\), \(\dfrac{IA}{ID}=1\)

\(\Rightarrow\dfrac{BD}{BC}.\dfrac{MC}{MA}.\dfrac{IA}{ID}=1\)

Theo định lý Menelaus đảo, suy ra B, I, M thẳng hàng.

Cách 2: Dùng vector

 Ta có \(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)

\(=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}.\dfrac{2}{3}\overrightarrow{BC}\)

\(=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\) 

\(=\dfrac{1}{6}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)

Lại có \(\overrightarrow{BM}=\dfrac{MC}{AC}\overrightarrow{BA}+\dfrac{MA}{AC}\overrightarrow{BC}\)

\(=\dfrac{3}{5}\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\)

\(=\dfrac{1}{5}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)

\(=\dfrac{6}{5}.\dfrac{1}{6}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)

\(=\dfrac{6}{5}\overrightarrow{BI}\)

Vậy \(\overrightarrow{BM}=\dfrac{6}{5}\overrightarrow{BI}\), suy ra B, I, M thẳng hàng.