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Xét ΔBAD có BM là đường trung tuyến
nên \(\overrightarrow{BM}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{5}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{6}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)\)
\(=\dfrac{5}{6}\left(\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\right)\)
\(\overrightarrow{BN}=\overrightarrow{BA}+\overrightarrow{AN}\)
\(=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\)
=>\(\overrightarrow{BM}=\dfrac{5}{6}\cdot\overrightarrow{BN}\)
=>B,M,N thẳng hàng
\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AD}\)(D là trung điểm của BC) (1)
\(\overrightarrow{AM}+\overrightarrow{AN}=2\overrightarrow{AK}\)(K là trung điểm của MN) (2)
Lấy (1) trừ (2) có: \(\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=2\left(\overrightarrow{AD}-\overrightarrow{AK}\right)\)
⇔\(\dfrac{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\overrightarrow{AM}+\overrightarrow{AN}\right)}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\right)}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\overrightarrow{AB}+\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\dfrac{1}{2}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)=\(\overrightarrow{KD}\)
a: \(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}\)
\(=\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\overrightarrow{BA}-\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
\(=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
Ta có:
\(\vec{AN}=\vec{AM}+\vec{MN}\)
\(=\dfrac{2}{3}\vec{AC}+\dfrac{1}{4}\vec{MB}\)
\(=\dfrac{2}{3}\vec{AC}+\dfrac{1}{4}\left(\vec{AB}-\vec{AM}\right)\)
\(=\dfrac{1}{4}\vec{AB}+\dfrac{1}{2}\vec{AC}\)
\(\vec{AP}=\vec{AC}+\vec{CP}\)
\(=\vec{AC}+\dfrac{1}{k+1}\vec{CB}\)
\(=\vec{AC}+\dfrac{1}{k+1}\left(\vec{AB}-\vec{AC}\right)\)
\(=\dfrac{1}{k+1}\vec{AB}+\dfrac{k}{k+1}\vec{AC}\)
A, N, P thẳng hàng khi:
\(\dfrac{\dfrac{k}{k+1}}{\dfrac{1}{k+1}}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{4}}\Leftrightarrow k=2\)
Kết luận: \(k=2\)
\(\overrightarrow{KA}=-\overrightarrow{AK}=-\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=-\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)
\(=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{KD}=\overrightarrow{AD}-\overrightarrow{AK}=\overrightarrow{AD}+\overrightarrow{KA}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
M là trung điểm BC \(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
\(\overrightarrow{IM}=2\overrightarrow{AI}\Rightarrow\overrightarrow{IA}+\overrightarrow{AM}=2\overrightarrow{AI}\)
\(\Rightarrow-\overrightarrow{AI}+\overrightarrow{AM}=2\overrightarrow{AI}\)
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{3}\overrightarrow{AM}=\dfrac{1}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{BI}=\overrightarrow{BA}+\overrightarrow{AI}=-\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AC}=-\dfrac{5}{6}\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AC}\)
Đặt \(\overrightarrow{AK}=x.\overrightarrow{AC}\)
\(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}=-\overrightarrow{AB}+x.\overrightarrow{AC}\)
Do B, I, K thẳng hàng \(\Rightarrow\overrightarrow{BK}\) và \(BI\) cùng phương
\(\Rightarrow\dfrac{-1}{\left(-\dfrac{5}{6}\right)}=\dfrac{x}{\left(\dfrac{1}{6}\right)}\Rightarrow x=\dfrac{1}{5}\)
\(\Rightarrow\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{AC}=\dfrac{1}{5}\left(\overrightarrow{AK}+\overrightarrow{KC}\right)=\dfrac{1}{5}\overrightarrow{AK}+\dfrac{1}{5}\overrightarrow{KC}\)
\(\Rightarrow\dfrac{4}{5}\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{KC}\)
\(\Rightarrow4.\overrightarrow{AK}=1.\overrightarrow{KC}\Rightarrow4.\overrightarrow{KA}=1.\overrightarrow{CK}\)
\(\Rightarrow\left\{{}\begin{matrix}n=4\\m=1\end{matrix}\right.\)
Các kí hiệu em ghi như IM=2AI và nKA=mCK nó là đoạn thẳng hay có vecto?