\(\hept{\begin{cases}AM=NC\\AM||NC\end{cases}\Rightarrow NA||BC}\)

\(\Delta ABK\)có \(\hept{\begin{cases}MI||AK\\MA=MB\end{cases}\Rightarrow IB=IK}\)

\(\Delta CDI\)có \(\hept{\begin{cases}NK||IC\\ND=NC\end{cases}\Rightarrow KD=KI}\)

\(\Rightarrow DK=KI=IB\)

a) N trung điểm AD \(\Rightarrow AN=\frac{AD}{2}=\frac{BC}{2}\)

M trung điểm BC \(\Rightarrow MC=\frac{BC}{2}\Rightarrow AN=MC\)mà AN//MC

nên AMCN là hình bình hành \(\Rightarrow\overrightarrow{AM}=\overrightarrow{NC}\)

b) Tương tự câu a ta được \(\hept{\begin{cases}ND=BM=\frac{1}{2}BC\\ND//BM\end{cases}}\)=> NDMB là hình bình hành=> NB//DM (1)

Xét 2 tam giác ANI và NDK: \(\hept{\begin{cases}AN=ND=\frac{AD}{2}\\\widehat{NAI}=\widehat{DNK}\left(AM//NC\right)\\\widehat{ANI}=\widehat{NDK}\left(NB//MD\right)\end{cases}\Rightarrow\Delta ANI=\Delta NDK\left(g.c.g\right)}\)

\(\Rightarrow NI=DK\)(2)

(1), (2) => \(\overrightarrow{NI}=\overrightarrow{DK}\)

\(a,\overrightarrow{AB}-\overrightarrow{DA}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{0}=\overrightarrow{AD}\)

\(b,\overrightarrow{AM}=\dfrac{\overrightarrow{AO}+\overrightarrow{AB}}{2}=\dfrac{\overrightarrow{AB}}{2}+\dfrac{\dfrac{1}{2}\overrightarrow{AC}}{2}=\overrightarrow{\dfrac{AB}{2}}+\dfrac{1}{4}\overrightarrow{AC}\)

\(=\overrightarrow{\dfrac{AB}{2}}+\dfrac{\overrightarrow{AB}+\overrightarrow{BC}}{4}=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{\overrightarrow{BC}}{4}=\dfrac{1}{4}\overrightarrow{BC}+\dfrac{3}{4}\overrightarrow{AB}\left(1\right)\)

\(\overrightarrow{AN}=\overrightarrow{BN}-\overrightarrow{BA}=k.\overrightarrow{BC}+\overrightarrow{AB}\left(2\right)\)

\(\left(1\right)\left(2\right)A,M,N\) \(thẳng\) \(hàng\Leftrightarrow\dfrac{k}{\dfrac{1}{4}}=\dfrac{1}{\dfrac{3}{4}}\Leftrightarrow k=\dfrac{1}{3}\)

\(\overrightarrow{OC}-\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC}=\overrightarrow{BC}\)(1)

ABCD là hình bình hành

=>\(\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{BC}\)

=>\(\overrightarrow{BC}=\overrightarrow{BD}-\overrightarrow{BA}\left(2\right)\)

Từ (1) và (2) suy ra \(\overrightarrow{OC}-\overrightarrow{OB}=\overrightarrow{BD}-\overrightarrow{BA}\)

Do ABCD là hình bình hành

\(\Rightarrow\overrightarrow{AB}=\overrightarrow{DC}\)

\(\overrightarrow{BC}-\overrightarrow{AB}=\overrightarrow{BC}-\overrightarrow{DC}=\overrightarrow{BC}+\overrightarrow{CD}=\overrightarrow{BD}\)

bài 1

a CO-OB=BA

<=.> CO = BA +OB

<=> CO=OA ( LUÔN ĐÚNG )=>ĐPCM

b AB-BC=DB

<=> AB=DB+BC

<=> AB=DC(LUÔN ĐÚNG )=> ĐPCM

Cc DA-DB=OD-OC

<=> DA+BD= OD+CO

<=> BA= CD (LUÔN ĐÚNG )=> ĐPCM

d DA-DB+DC=0

VT= DA +BD+DC

= BA+DC

Mà BA=CD(CMT)

=> VT= CD+DC=O

BÀI 2

AC=AB+BC

BD=BA+AD

=> AC+BD= AB+BC+BA+AD=BC+AD (đpcm)

\(\overrightarrow{NC}=2\overrightarrow{ND}=2\overrightarrow{NC}+2\overrightarrow{CD}\Rightarrow\overrightarrow{NC}=2\overrightarrow{DC}\Rightarrow\overrightarrow{CN}=2\overrightarrow{CD}\)

a.

\(\overrightarrow{DM}=\overrightarrow{DC}+\overrightarrow{CM}=\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{CB}=\overrightarrow{AB}-\dfrac{1}{2}\overrightarrow{AD}\)

\(\overrightarrow{MN}=\overrightarrow{MC}+\overrightarrow{CN}=\dfrac{1}{2}\overrightarrow{BC}+2\overrightarrow{CD}=-2\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}\)

b.

\(\left\{{}\begin{matrix}\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}\\\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=-\overrightarrow{AB}+\overrightarrow{AD}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\dfrac{1}{2}\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{BD}\\\overrightarrow{AD}=\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BD}\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{MN}=-2\left(\dfrac{1}{2}\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{BD}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BD}\right)=-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{5}{4}\overrightarrow{BD}\)