Cho hình tứ diện ABCD. Chứng minh rằng:AB = 1/2AC + 1/2AD + 1/2CD + DB

Ta có: \(\frac{1}{2}\overrightarrow {AC} + \frac{1}{2}\overrightarrow {AD} + \frac{1}{2}\overrightarrow {CD} + \overrightarrow {DB} \) = \(\left( {\frac{1}{2}\overrightarrow {AC} + \frac{1}{2}\overrightarrow {CD} } \right)\) + \(\frac{1}{2}\overrightarrow {AD} \) + \(\overrightarrow {DB} \)

                                                        = \(\frac{1}{2}\overrightarrow {AD} + \frac{1}{2}\overrightarrow {AD} + \overrightarrow {DB} \)

                                                        = \(\overrightarrow {AD} + \overrightarrow {DB} \)

                                                        = \(\overrightarrow {AB} \).

Vậy \(\overrightarrow {AB} = \frac{1}{2}\overrightarrow {AC} + \frac{1}{2}\overrightarrow {AD} + \frac{1}{2}\overrightarrow {CD} + \overrightarrow {DB} \).