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\(\overrightarrow{AD}+2\overrightarrow{AB}=\overrightarrow{AD}+\overrightarrow{AB}+\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{AB}=2\overrightarrow{AI}\) (đpcm)
a: vecto BM=vecto BA+vecto AM
=-vecto AB+1/2vecto AD
vecto AN=vecto AD+vecto DN
=vecto AD+1/2*vecto AB
b: vecto BM*vecto AN=vecto 0
=>BM vuông góc với AN
Do M là trung điểm BC nên: \(\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
Tương tự: \(\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{BC}\) ; \(\overrightarrow{CP}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)
Cộng vế:
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BA}\right)+\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)+\dfrac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CB}\right)=\overrightarrow{0}\)
b. Từ câu a ta có:
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{AO}+\overrightarrow{OM}+\overrightarrow{BO}+\overrightarrow{ON}+\overrightarrow{CO}+\overrightarrow{OP}=\overrightarrow{0}\)
\(\Leftrightarrow-\overrightarrow{OA}+\overrightarrow{OM}-\overrightarrow{OB}+\overrightarrow{ON}-\overrightarrow{OC}+\overrightarrow{OP}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OM}+\overrightarrow{ON}+\overrightarrow{OP}\) (đpcm)
\(\overrightarrow{AC}+\overrightarrow{BM}=\left(\overrightarrow{AM}+\overrightarrow{MC}\right)+\left(\overrightarrow{BC}+\overrightarrow{CM}\right)=\overrightarrow{AM}+\overrightarrow{BC}+\left(\overrightarrow{MC}+\overrightarrow{CM}\right)=\overrightarrow{AM}+\overrightarrow{BC}\)
Xét ΔABC có AD là phân giác
nên \(\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{5}{7}\)
=>\(\dfrac{BD}{5}=\dfrac{DC}{7}\)
mà BD+DC=BC=6
nên \(\dfrac{BD}{5}=\dfrac{CD}{7}=\dfrac{BD+CD}{5+7}=\dfrac{6}{12}=\dfrac{1}{2}\)
=>BD=2,5; CD=3,5
=>\(\dfrac{BD}{BC}=\dfrac{5}{12};\dfrac{CD}{CB}=\dfrac{7}{12}\)
\(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\)
\(=\overrightarrow{AB}+\dfrac{5}{12}\cdot\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{5}{12}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{7}{12}\cdot\overrightarrow{AB}+\dfrac{5}{12}\cdot\overrightarrow{AC}\)
=>Chọn C
Lời giải:
a)
$2\overrightarrow{AD}=\overrightarrow{AD}+\overrightarrow{AD}$
$=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{AC}+\overrightarrow{CD}$
$=\overrightarrow{AB}+\overrightarrow{AC}+(\overrightarrow{BD}+\overrightarrow{CD})$
$=\overrightarrow{AB}+\overrightarrow{AC}$
$\Rightarrow \overrightarrow{AD}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}$
Tương tự:
$\overrightarrow{BE}=\frac{\overrightarrow{BC}+\overrightarrow{BA}}{2}$
$\overrightarrow{CF}=\frac{\overrightarrow{CA}+\overrightarrow{CB}}{2}$
Cộng lại:
$\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\frac{\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{AC}+\overrightarrow{CA}+\overrightarrow{BC}+\overrightarrow{CB}}{2}=\frac{\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}}{2}=\overrightarrow{0$}$
Ta có đpcm.
b)
$\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{MD}+\overrightarrow{DA}+\overrightarrow{ME}+\overrightarrow{EB}+\overrightarrow{MF}+\overrightarrow{FC}$
$=(\overrightarrow{MD}+\overrightarrow{ME}+\overrightarrow{MF})+(\overrightarrow{DA}+\overrightarrow{EB}+\overrightarrow{FC})$
$=(\overrightarrow{MD}+\overrightarrow{ME}+\overrightarrow{MF})-(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF})$
$=\overrightarrow{MD}+\overrightarrow{ME}+\overrightarrow{MF}-\overrightarrow{0}$ (theo phần a)
$=\overrightarrow{MD}+\overrightarrow{ME}+\overrightarrow{MF}$
Ta có đpcm.
Vì ABCD là hình chữ nhật
nên \(\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC};\overrightarrow{BA}+\overrightarrow{BC}=\overrightarrow{BD}\)
=>\(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=\left|\overrightarrow{AC}\right|=AC;\left|\overrightarrow{BA}+\overrightarrow{BC}\right|=\left|\overrightarrow{BD}\right|=BD\)
mà AC=BD(ABCD là hình chữ nhật)
nên \(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=\left|\overrightarrow{BA}+\overrightarrow{BC}\right|\)