$AB=a;AD=a\sqrt{2}$

$K$ là trung điểm $AD$

`=>\vec{BA}+\vec{BD}=2\vec{BK}`

`=>\vec{BK}=(\vec{BA}+\vec{BD})/2={\vec{BA}+\vec{BA}+\vec{AD}}/2`

`=>\vec{BK}=(2\vec{BA}+\vec{AD})/2={2\vec{BA}+\vec{BC}}/2`

Ta có:

`\vec{AC}=\vec{AD}+\vec{DC}=\vec{AD}+\vec{AB}`

`=>2\vec{AC}.\vec{BK}=(vec{AD}+\vec{AB}).(2\vec{BA}+\vec{BC})`

`=2\vec{AD}.\vec{BA}+2\vec{AB}.\vec{BA}+ \vec{AD}.\vec{BC}+ \vec{AB}.\vec{BC}`

`=0+2\vec{AB}.(-\vec{AB})+\vec{AD}.\vec{BC}+ 0`

`=-2(\vec{AB})^2+\vec{AD}.\vec{AD}`

`=-2AB^2+AD^2`

`=-2a^2+(a\sqrt{2})^2=-2a^2+2a^2=0`

`=>\vec{BK}`$\perp \vec{AC}$

_______.

Cho nên:

`2\vec{AB}.\vec{BA}=2\vec{AB}.(-\vec{AB})=-2(\vec{AB})^2=-2AB^2`

`\vec{AD}.\vec{BC}=\vec{AD}.\vec{AD}=(\vec{AD})^2=AD^2`

Còn phần sau cách viết "Thay AB = vecto AD.BC= (vectoAD)^2=AD^2" không chính xác

AB=a;AD=a√2AB=a;AD=a2

KK là trung điểm ADAD

⇒−−→BA+−−→BD=2−−→BK⇒BA→+BD→=2BK→

⇒−−→BK=−−→BA+−−→BD2=−−→BA+−−→BA+−−→AD2⇒BK→=BA→+BD→2=BA→+BA→+AD→2

⇒−−→BK=2−−→BA+−−→AD2=2−−→BA+−−→BC2⇒BK→=2BA→+AD→2=2BA→+BC→2

Ta có:

−−→AC=−−→AD+−−→DC=−−→AD+−−→ABAC→=AD→+DC→=AD→+AB→

⇒2−−→AC.−−→BK=(−−→AD+−−→AB).(2−−→BA+−−→BC)⇒2AC→.BK→=(AD→+AB→).(2BA→+BC→)

=2−−→AD.−−→BA+2−−→AB.−−→BA+−−→AD.−−→BC+−−→AB.−−→BC=2AD→.BA→+2AB→.BA→+AD→.BC→+AB→.BC→

=0+2−−→AB.(−−−→AB)+−−→AD.−−→BC+0=0+2AB→.(-AB→)+AD→.BC→+0

=−2(−−→AB)2+−−→AD.−−→AD=-2(AB→)2+AD→.AD→

=−2AB2+AD2=-2AB2+AD2

=−2a2+(a√2)2=−2a2+2a2=0=-2a2+(a2)2=-2a2+2a2=0

⇒−−→BK⇒BK→⊥→AC⊥AC→

_______.

Cho nên:

2−−→AB.−−→BA=2−−→AB.(−−−→AB)=−2(−−→AB)2=−2AB22AB→.BA→=2AB→.(-AB→)=-2(AB→)2=-2AB2

−−→AD.−−→BC=−−→AD.−−→AD=(−−→AD)2=AD2