Question

Cho hình bình hành ABCD. Gọi I, J lần lượt là trung điểm BC và CD

a) Chứng minh rằng: \(\overrightarrow{MA}\) + \(\overrightarrow{MC}\) = \(\overrightarrow{MB}\) + \(\overrightarrow{MD}\) với mọi M

b) Chứng minh rằng: 2 ( \(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{JA}+\overrightarrow{DA}\) ) = 3\(\overrightarrow{DB}\)

c) Trên BC lấy điểm H, trên BD lấy điểm K sao cho \(\overrightarrow{BH}\) = \(\frac{1}{5}\overrightarrow{BC}\), \(\overrightarrow{BK}=\frac{1}{6}\overrightarrow{BD}\). Chứng minh rằng A, H, K thẳng hàng

Answer 1

\(\overrightarrow{MA}+\overrightarrow{MC}=\overrightarrow{MB}+\overrightarrow{BA}+\overrightarrow{MD}+\overrightarrow{DC}=\overrightarrow{MB}+\overrightarrow{MD}+\overrightarrow{BA}+\overrightarrow{DC}=\overrightarrow{MB}+\overrightarrow{MD}\)

b/

\(2\left(\overrightarrow{JA}+\overrightarrow{AB}+\overrightarrow{DA}+\overrightarrow{AI}\right)=2\left(\overrightarrow{JB}+\overrightarrow{DI}\right)=2\left(\overrightarrow{JD}+\overrightarrow{DB}+\overrightarrow{DB}+\overrightarrow{BI}\right)\)

\(=2\left(2\overrightarrow{DB}+\overrightarrow{IC}+\overrightarrow{CJ}\right)=2\left(2\overrightarrow{DB}+\overrightarrow{IJ}\right)=2\left(2\overrightarrow{DB}+\frac{1}{2}\overrightarrow{BD}\right)=3\overrightarrow{DB}\)c/

\(\overrightarrow{AK}=\overrightarrow{AB}+\overrightarrow{BK}=\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BD}=\overrightarrow{AB}+\frac{1}{6}\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=\frac{5}{6}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BC}\)

\(\overrightarrow{AH}=\overrightarrow{AB}+\overrightarrow{BH}=\overrightarrow{AB}+\frac{1}{5}\overrightarrow{BC}=\frac{6}{5}\left(\frac{5}{6}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BC}\right)=\frac{6}{5}\overrightarrow{AK}\)

\(\Rightarrow A;K;H\) thẳng hàng