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a) Vì M, N, P lần lượt là trung điểm của BC, CA, AB
Nên AM, BN, CP lần lượt là đường trung tuyến của BC, CA, AB.
\(\Rightarrow\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\overrightarrow{0}\)
Lời giải:
a)
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{BC}+\overrightarrow{CN}+\overrightarrow{CA}+\overrightarrow{AP}\)
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\overrightarrow{AC}+\overrightarrow{CM}+\overrightarrow{BA}+\overrightarrow{AN}+\overrightarrow{CB}+\overrightarrow{BP}\)
\(\Rightarrow 2(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP})=(\overrightarrow{AB}+\overrightarrow{BA})+(\overrightarrow{BM}+\overrightarrow{CM})+(\overrightarrow{BC}+\overrightarrow{CB})+(\overrightarrow{CA}+\overrightarrow{AC})+(\overrightarrow{AP}+\overrightarrow{BP})+(\overrightarrow{CN}+\overrightarrow{AN})\)
\(=\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\) (do các cặp tổng đều là vecto đối nhau)
\(\Rightarrow \overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=0\)
(đpcm)
b) Theo phần a:
\(\overrightarrow{AM}=-(\overrightarrow{BN}+\overrightarrow{CP})=-\overrightarrow{BN}+(-\overrightarrow{CP})\)
\(=\overrightarrow{NB}+\overrightarrow{PC}\) (đpcm)
Bài 1 và Bài 2 tương tự nhau nên mk sẽ chỉ CM bài 1 thôi nha
Có \(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\overrightarrow{AB}+\overrightarrow{CD}=0\)
\(\Rightarrow\overrightarrow{AD}+\overrightarrow{DB}+\overrightarrow{CB}+\overrightarrow{BD}=0\)
\(\Leftrightarrow\overrightarrow{AD}+\overrightarrow{CB}=0\Leftrightarrow\overrightarrow{AD}=\overrightarrow{BC}\)
Bài 3:
Xét \(\Delta AIP\) theo quy tắc trung điểm có:
\(\overrightarrow{IC}=\frac{\overrightarrow{IA}+\overrightarrow{IP}}{2}\)
Làm tương tự vs các tam giác còn lại
\(\Rightarrow\overrightarrow{IB}=\frac{\overrightarrow{IN}+\overrightarrow{IC}}{2}\)
\(\Rightarrow\overrightarrow{IA}=\frac{\overrightarrow{IB}+\overrightarrow{IM}}{2}\)
Cộng vế vs vế
\(\Rightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\frac{\overrightarrow{IA}+\overrightarrow{IP}+\overrightarrow{IN}+\overrightarrow{IC}+\overrightarrow{IB}+\overrightarrow{IM}}{2}\)
\(\Leftrightarrow2\overrightarrow{IA}+2\overrightarrow{IB}+2\overrightarrow{IC}=\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{IM}+\overrightarrow{IN}+\overrightarrow{IP}\)
\(\Leftrightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{IM}+\overrightarrow{IN}+\overrightarrow{IP}\left(đpcm\right)\)
a: \(\overrightarrow{AM}+\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}=\dfrac{1}{2}\overrightarrow{AC}\)
b: \(=\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
\(=\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
c: \(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{CA}\)
\(=\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)=\overrightarrow{0}\)
a) Chữa đề: \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)
\(Ta\text{ }có:\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{DA}+\overrightarrow{AB}\\ =\overrightarrow{CB}+\overrightarrow{DA}+\left(\overrightarrow{BA}+\overrightarrow{AB}\right)=\overrightarrow{CB}+\overrightarrow{DA}\)
\(\)\(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CA}+\overrightarrow{CB}+\overrightarrow{DC}\\ =2\overrightarrow{CM}+2\overrightarrow{NC}=2\left(\overrightarrow{NC}+\overrightarrow{CM}\right)=2\overrightarrow{NM}\)
Vậy \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)
\(\text{b) }\overrightarrow{AD}+\overrightarrow{BD}+\overrightarrow{AC}+\overrightarrow{BC}=-\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{CA}+\overrightarrow{CB}\right)\\ =-\left[\left(\overrightarrow{DA}+\overrightarrow{DB}\right)+\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\right]\\ =-\left(2\overrightarrow{DM}+2\overrightarrow{CM}\right)=2\left(\overrightarrow{MD}+\overrightarrow{MC}\right)=4\left(\overrightarrow{MN}\right)\)
\(\text{c) }2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{DA}\right)+\left(\overrightarrow{AI}+\overrightarrow{NA}\right)\right]\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{DB}\right)+\overrightarrow{NI}\right]=2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)\)
Mà IN là dường trung bình \(\Delta BCD\)
\(\Rightarrow\left\{{}\begin{matrix}IN//BD\\IN=\frac{1}{2}BD\end{matrix}\right.\Rightarrow\overrightarrow{IN}=\frac{1}{2}\overrightarrow{BD}\\ \Rightarrow2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)=2\left(\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DB}\right)=2\cdot\frac{3}{2}\overrightarrow{DB}=3\overrightarrow{DB}\)
a/ \(\overrightarrow{AN}+\overrightarrow{BP}+\overrightarrow{CM}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)+\frac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{BA}\right)+\frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\)
\(=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BA}\right)+\frac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)+\frac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CB}\right)=\overrightarrow{0}\)
b/
Do MN là đường trung bình tam giác ABC \(\Rightarrow\overrightarrow{MN}=\frac{1}{2}\overrightarrow{AC}\)
\(\overrightarrow{AN}=\overrightarrow{AM}+\overrightarrow{MN}=\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AC}=\overrightarrow{AM}+\overrightarrow{AP}\)
c/
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}+\frac{1}{2}\overrightarrow{CA}=\frac{1}{2}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CA}=\overrightarrow{0}\)
ta có : 2vecto AM =vecto AB +vectoAC
2 vecto BN = vectoBA +vectoBC
2 vecto CP =vecto CA + vectoCB
=> 2vecto AM +2 vecto BN + 2 vecto CP =vecto AB +vecto AC +vecto BA +vecto BC +vecto CA + vecto CB =vecto 0
=> vecto AM + vecto BN + vecto CP =vecto 0
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