a: \(\overrightarrow{CA}+\overrightarrow{AB}+\overrightarrow{BC}\)

\(=\overrightarrow{CB}+\overrightarrow{BC}\)

\(=\overrightarrow{0}\)

b: \(\overrightarrow{AM}+\overrightarrow{AP}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\dfrac{1}{2}\cdot2\cdot\overrightarrow{AN}=\overrightarrow{AN}\)

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. 

a: \(\overrightarrow{BM}=\overrightarrow{BA}+\overrightarrow{AM}=\overrightarrow{BA}+\frac12\cdot\overrightarrow{AD}\)

\(=\overrightarrow{BA}+\frac12\cdot\frac12\cdot\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=-\frac34\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)

\(=-\frac14\left(3\cdot\overrightarrow{AB}-\overrightarrow{AC}\right)\)

\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}=\overrightarrow{BA}+\frac13\cdot\overrightarrow{AC}=\frac{-1}{3}\left(3\cdot\overrightarrow{AB}-\overrightarrow{AC}\right)\)

=>\(\frac{\overrightarrow{BM}}{\overrightarrow{BN}}=\frac{-1}{4}:\frac{-1}{3}=\frac34\)

=>B.M,N thẳng hàng

b: \(\overrightarrow{IM}=\overrightarrow{IA}+\overrightarrow{AM}\)

\(=-\frac23\cdot\overrightarrow{AB}+\frac12\cdot\overrightarrow{AD}=-\frac23\cdot\overrightarrow{AB}+\frac12\cdot\frac12\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

\(=-\frac23\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}=\frac{-5}{12}\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)

\(=\frac14\left(-\frac53\cdot\overrightarrow{AB}+\overrightarrow{AC}\right)\)

\(\overrightarrow{IJ}=\overrightarrow{IA}+\overrightarrow{AJ}=-\frac23\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{AC}=-2\left(\frac13\cdot\overrightarrow{AB}-\frac15\cdot\overrightarrow{AC}\right)\)

\(=\frac25\left(\frac{-5}{3}\cdot\overrightarrow{AB}+\overrightarrow{AC}\right)\)

Do đó: \(\frac{\overrightarrow{IM}}{\overrightarrow{IJ}}=\frac14:\frac25=\frac58\)

=>I,M,J thẳng hàng

a; Xét ΔCAB có

M,N lần lượt là trung điểm của CB,CA

=>MN là đường trung bình của ΔCAB

=>MN//AB và \(MN=\frac{AB}{2}\)

MN//AB

=>MN//AP và MN//BP

Ta có: \(MN=\frac{AB}{2}\)

\(AP=BP=\frac{AB}{2}\)

Do đó: MN=AP=BP

Xét tứ giác MNAP có

MN//AP

MN=AP

Do đó: MNAP là hình bình hành

=>\(\overrightarrow{MN}=\overrightarrow{PA}\)

mà \(\overrightarrow{BP}=\overrightarrow{PA}\) (P là trung điểm của BA)

nên \(\overrightarrow{MN}=\overrightarrow{BP}\)

\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AE}\)

\(\overrightarrow{AM}+\overrightarrow{AN}=2\overrightarrow{AE}\)

\(\Rightarrow\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AM}+\overrightarrow{AN}\)

a: \(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)