Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ta có : \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AM}+\overrightarrow{MN}+\overrightarrow{NB}+\overrightarrow{DM}+\overrightarrow{MN}+\overrightarrow{NC}\)
\(=2\overrightarrow{MN}+\left(\overrightarrow{AM}+\overrightarrow{DM}\right)+\left(\overrightarrow{NB}+\overrightarrow{NC}\right)=2\overrightarrow{MN}\left(đpcm\right)\)
b) ta có : \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AI}+\overrightarrow{IJ}+\overrightarrow{JB}+\overrightarrow{CI}+\overrightarrow{IJ}+\overrightarrow{JD}\)
\(=2\overrightarrow{IJ}+\left(\overrightarrow{AI}+\overrightarrow{CI}\right)+\left(\overrightarrow{JB}+\overrightarrow{JD}\right)=2\overrightarrow{IJ}\left(đpcm\right)\)
bn dùng định lí ta lét chứng minh được \(\overrightarrow{MJ}=\overrightarrow{IN}=\dfrac{1}{2}\overrightarrow{AB}\)
C) ta có : \(\overrightarrow{MN}+\overrightarrow{IJ}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{IA}+\overrightarrow{AB}+\overrightarrow{BJ}\)
\(=2\overrightarrow{AB}+\left(\overrightarrow{MA}+\overrightarrow{BJ}\right)+\left(\overrightarrow{BN}+\overrightarrow{IA}\right)\)
\(=2\overrightarrow{AB}+\left(\overrightarrow{DM}+\overrightarrow{JD}\right)+\left(\overrightarrow{NC}+\overrightarrow{CI}\right)=2\overrightarrow{AB}+\overrightarrow{JM}+\overrightarrow{NI}\) \(=2\overrightarrow{AB}+\overrightarrow{BA}=\overrightarrow{AB}\left(đpcm\right)\)d) ta có : \(\overrightarrow{IM}+\overrightarrow{IN}=\overrightarrow{IJ}+\overrightarrow{JM}+\overrightarrow{IN}=\overrightarrow{IJ}\left(đpcm\right)\)
a)Ta có:
\(\overrightarrow{OA}+\overrightarrow{OM}+\overrightarrow{ON}=\overrightarrow{CO}+\dfrac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OC}+\overrightarrow{OD}\right)\)
\(=\overrightarrow{CO}+\dfrac{1}{2}.2\overrightarrow{OC}\)
\(=\overrightarrow{0}\)
\(\RightarrowĐPCM\)
b) Ta có:
\(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+2\overrightarrow{AB}\right)\)
\(\Rightarrow2\overrightarrow{AM}=\overrightarrow{AD}+2\overrightarrow{AB}\) (1)
Mà \(2\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{AC}\)(2)
Từ (1)(2) =>\(\overrightarrow{AD}+2\overrightarrow{AB}=\overrightarrow{AB}+\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AC}+\overrightarrow{AB}=\overrightarrow{AB}+\overrightarrow{AC}\)
\(\RightarrowĐPCM\)
A B C D O M N E F
a) Giả sử \(\overrightarrow{OA}+\overrightarrow{OC}=\overrightarrow{OB}+\overrightarrow{OD}\)
\(\Leftrightarrow\overrightarrow{OA}+\overrightarrow{OC}-\overrightarrow{OB}-\overrightarrow{OD}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{OA}+\overrightarrow{BO}+\overrightarrow{OC}+\overrightarrow{DO}=\overrightarrow{0}\)
\(\Leftrightarrow\left(\overrightarrow{BO}+\overrightarrow{OA}\right)+\left(\overrightarrow{DO}+\overrightarrow{OC}\right)=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{BA}+\overrightarrow{DC}=\overrightarrow{0}\) (đúng do tứ giác ABCD là hình bình hành).
b) \(\overrightarrow{ME}+\overrightarrow{FN}=\overrightarrow{MA}+\overrightarrow{AE}+\overrightarrow{FC}+\overrightarrow{CN}\)
\(=\left(\overrightarrow{MA}+\overrightarrow{CN}\right)+\left(\overrightarrow{AE}+\overrightarrow{FC}\right)\).
Do các tứ giác AMOE, MOFB, OFCN, EOND cũng là các hình bình hành.
Vì vậy \(\overrightarrow{CN}=\overrightarrow{FO}=\overrightarrow{BM};\overrightarrow{FC}=\overrightarrow{ON}=\overrightarrow{ED}\).
Do đó: \(\overrightarrow{ME}+\overrightarrow{FN}=\left(\overrightarrow{MA}+\overrightarrow{CN}\right)+\left(\overrightarrow{AE}+\overrightarrow{FC}\right)\)
\(=\left(\overrightarrow{MA}+\overrightarrow{BM}\right)+\left(\overrightarrow{AE}+\overrightarrow{ED}\right)\)
\(=\overrightarrow{BA}+\overrightarrow{AD}=\overrightarrow{BD}\) (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
\(\Rightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\frac{\overrightarrow{IA}+\overrightarrow{IP}+\overrightarrow{IN}+\overrightarrow{IC}+\overrightarrow{IB}+\overrightarrow{IM}}{2}\)
\(\Leftrightarrow2\overrightarrow{IA}+2\overrightarrow{IB}+2\overrightarrow{IC}=\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{IM}+\overrightarrow{IN}+\overrightarrow{IP}\)
\(\Leftrightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{IM}+\overrightarrow{IN}+\overrightarrow{IP}\left(đpcm\right)\)