Em ms hok cái này nên ko chắc lăm ạ :D

Theo quy tắc 3 điểm\(\Rightarrow\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{BC}\)

\(\overrightarrow{AM}+\overrightarrow{AN}=\overrightarrow{MN}\)

Có I là TĐ của BC\(\Rightarrow\overrightarrow{EB}+\overrightarrow{EC}=\overrightarrow{BC}=0\) (1)

Có I là TĐ của MN \(\Rightarrow\overrightarrow{EM}+\overrightarrow{EN}=\overrightarrow{MN}=0\) (2)

Từ (1) và (2)\(\Rightarrowđpcm\)

a) Ta có: \(\overrightarrow {BC} ,\overrightarrow {PN} \) là hai vecto cùng hướng và \(\frac{1}{2}\left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {PN} } \right|\)

\( \Rightarrow \frac{1}{2}\overrightarrow {BC}  = \overrightarrow {PN} \)\( \Rightarrow \overrightarrow {AP}  + \frac{1}{2}\overrightarrow {BC}  = \overrightarrow {AP}  + \overrightarrow {PN}  = \overrightarrow {AN} \)

b) Ta có: \(\overrightarrow {MP} ,\overrightarrow {CA} \) là hai vecto cùng hướng và \(2\left| {\overrightarrow {MP} } \right| = \left| {\overrightarrow {CA} } \right|\)

\( \Rightarrow 2\overrightarrow {MP}  = \overrightarrow {CA} \)\( \Rightarrow \overrightarrow {BC}  + 2\overrightarrow {MP}  = \overrightarrow {BC}  + \overrightarrow {CA}  = \overrightarrow {BA} \)

a)

\(=\frac{1}{2}\left(\overrightarrow{BA}+\frac{1}{2}\overrightarrow{BC}\right)\)

\(=\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}\)

\(=\overrightarrow{BA}+\frac{1}{3}\overrightarrow{AC}\)

\(=\overrightarrow{BA}+\frac{1}{3}\left(\overrightarrow{BC}-\overrightarrow{BA}\right)\)

\(=\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}-\frac{1}{3}\overrightarrow{BA}\)

\(=\frac{2}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}\)

b) Ta có: \(\overrightarrow{BK}=\frac{2}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}=\frac{4}{3}\left(\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}\right)=\frac{4}{3}\overrightarrow{BI}\)

=> B,K,I thẳng hàng

c) \(27\overrightarrow{MA}-8\overrightarrow{MB}=2015\overrightarrow{MC}\)

\(\Leftrightarrow27\left(\overrightarrow{MC}+\overrightarrow{CA}\right)-8\left(\overrightarrow{MC}+\overrightarrow{CB}\right)=2015\overrightarrow{MC}\)

\(\Leftrightarrow27\overrightarrow{MC}+27\overrightarrow{CA}-8\overrightarrow{MC}-8\overrightarrow{CB}-2015\overrightarrow{MC}=\overrightarrow{0}\)

\(\Leftrightarrow-1996\overrightarrow{MC}+27\overrightarrow{CA}-8\overrightarrow{CB}=\overrightarrow{0}\)

\(\Leftrightarrow1996\overrightarrow{CM}=8\overrightarrow{CB}-27\overrightarrow{CA}\)

\(\Leftrightarrow\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)

Vậy: Dựng điểm M sao cho \(\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)

Có vẻ không đúng.

Giả sử \(\overrightarrow{AB}+\overrightarrow{MB}+\overrightarrow{MA}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{MB}+\left(\overrightarrow{MA}+\overrightarrow{AB}\right)=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{MB}+\overrightarrow{MB}=\overrightarrow{0}\)

\(\Leftrightarrow2\overrightarrow{MB}=\overrightarrow{0}\)

\(\Leftrightarrow M\equiv B\) (Vô lí)

Hình vẽ:

a, Chứng minh \(\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{MA}=\overrightarrow{0}\)

Ta có \(\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{MA}=\overrightarrow{BM}+\left(\overrightarrow{MA}+\overrightarrow{AB}\right)=\overrightarrow{BM}+\overrightarrow{MB}=\overrightarrow{0}\)

b, Gọi H là trung điểm \(MC\)

Ta có \(AM=\sqrt{AC^2-MC^2}=\sqrt{4a^2-a^2}=a\sqrt{3}\)

\(AH=\sqrt{AM^2+MH^2}=\sqrt{\left(a\sqrt{3}\right)^2+\left(\dfrac{a}{2}\right)^2}=a.\dfrac{\sqrt{13}}{2}\)

\(\left|\overrightarrow{AM}+\overrightarrow{AC}\right|=\left|2\overrightarrow{AH}\right|=2AH=a\sqrt{13}\)

c, Gọi D là trung điểm AB

\(3\overrightarrow{NA}+3\overrightarrow{NB}+2\overrightarrow{NC}=3\left(\overrightarrow{NA}+\overrightarrow{NB}\right)+2\overrightarrow{NC}=6\overrightarrow{ND}+2\overrightarrow{NC}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{NC}=3\overrightarrow{DN}\)

Vậy N thuộc đoạn CD sao cho \(CN=\dfrac{3}{4}CD\)

a/ \(\overrightarrow{AC}=3\overrightarrow{AM};\overrightarrow{BN}=\frac{1}{2}\overrightarrow{BC}\)

\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}=\frac{1}{3}\overrightarrow{CA}+\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}\)

\(=\frac{1}{3}\overrightarrow{CB}+\frac{1}{3}\overrightarrow{CD}+\overrightarrow{DC}+\frac{1}{2}\overrightarrow{BC}=\frac{2}{3}\overrightarrow{DC}+\frac{1}{6}\overrightarrow{BC}=\frac{2}{3}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BA}+\frac{1}{6}\overrightarrow{AC}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\)

Hmm, MN làm sao vuông góc vs BC đc. Nó chỉ vuông góc khi M là TĐ của AC thôi, bởi N là trung điểm của BC rồi mà, hại não :((

2/\(\overrightarrow{BK}=\frac{4}{13}\overrightarrow{BA}\Rightarrow\overrightarrow{BC}+\overrightarrow{CK}=\frac{4}{13}\overrightarrow{BC}+\frac{4}{13}\overrightarrow{CA}\)

\(\Leftrightarrow\overrightarrow{CK}=\frac{9}{13}\overrightarrow{CB}+\frac{4}{13}\overrightarrow{CA}\)

\(\overrightarrow{GB}+\overrightarrow{GM}+\overrightarrow{GN}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{GC}+\overrightarrow{CB}+\overrightarrow{GC}+\overrightarrow{CM}+\overrightarrow{GC}+\overrightarrow{CN}=\overrightarrow{0}\)

\(\Leftrightarrow3\overrightarrow{GC}+\overrightarrow{CB}+\overrightarrow{CN}+\overrightarrow{NM}+\overrightarrow{CN}=\overrightarrow{0}\)

\(\Leftrightarrow3\overrightarrow{GC}+\overrightarrow{CB}+2\overrightarrow{CN}+\frac{1}{2}\overrightarrow{BA}+\frac{1}{6}\overrightarrow{CA}=\overrightarrow{0}\)

Ta có : \(\overrightarrow{CN}=\frac{1}{2}\overrightarrow{CB}\Rightarrow3\overrightarrow{GC}+\overrightarrow{CB}+\overrightarrow{CB}+\frac{1}{2}\overrightarrow{BA}+\frac{1}{6}\overrightarrow{CA}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{CG}=\frac{2}{3}\overrightarrow{CB}+\frac{1}{6}\overrightarrow{BA}+\frac{1}{18}\overrightarrow{CA}\)

\(\Leftrightarrow\overrightarrow{CG}=\frac{2}{3}\overrightarrow{CB}+\frac{1}{6}\overrightarrow{BC}+\frac{1}{6}\overrightarrow{CA}+\frac{1}{18}\overrightarrow{CA}\)

\(=\frac{1}{2}\overrightarrow{CB}+\frac{2}{9}\overrightarrow{CA}\)

Có \(\overrightarrow{CK}=\frac{18}{13}\overrightarrow{CG}\Rightarrow\) C,G,K thẳng hàng