Cho hình thang cân ABCD có CD = 2AB = 2a,(a > 0), \(\widehat {DAB}\) = 120°, AH vuông góc CD tại H. Tính \(\overrightarrow {AH} \left( {\overrightarrow {CD} - 4\overrightarrow {AD} } \right),\overrightarrow {AC} .\overrightarrow {BH} \).

Ta có: \(\widehat {DAB}\) = 120° suy ra: \(\widehat {DAH}\) = 30°

Vì ABCD là hình thang cân nên DH = (CD – AB) : 2 = (2a – a) : 2 = \(\frac{a}{2}\)

Xét tam giác vuông ADH ta có:

AD = \(\frac{{AH}}{{\sin 30^\circ }} = \frac{a}{{2.\frac{1}{2}a}} = a\)

AH = \(\sqrt {A{D^2} - D{H^2}} = \sqrt {{a^2} - \frac{{{a^2}}}{4}} = \frac{{a\sqrt 3 }}{2}\)

\(\overrightarrow {AH} \left( {\overrightarrow {CD} - 4\overrightarrow {AD} } \right) = \overrightarrow {AH} .\overrightarrow {CD} - 4\overrightarrow {AH} .\overrightarrow {AD} \)

\[ = AH.CD.\cos \left( {\overrightarrow {AH} ,\overrightarrow {CD} } \right) - 4.AH.AD.\cos \left( {\overrightarrow {AH} ,\overrightarrow {AD} } \right)\]

\[ = AH.CD.\cos 90^\circ - 4.AH.AD.\cos 30^\circ \]

\[ = - 4.\frac{{a\sqrt 3 }}{2}.a.\frac{{\sqrt 3 }}{2} = - 3{a^2}\]

+) \(\overrightarrow {AC} .\overrightarrow {BH} = \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right)\left( {\overrightarrow {BC} + \overrightarrow {CH} } \right)\)

\(\overrightarrow {AC} .\overrightarrow {BH} = \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right)\left( {\overrightarrow {BC} + \overrightarrow {CH} } \right)\)

\( = AB.BC.\cos 60^\circ + AB.CH.\cos 90^\circ + {\rm B}{C^2} + BC.CH.\cos 120^\circ \)

\( = a.a.\frac{1}{2} + {a^2} + a.\frac{{3a}}{2}.\left( {\frac{{ - 1}}{2}} \right) = \frac{{3{a^2}}}{4}\).