Kẻ $DH\bot
AB, CM// AD$
$\Rightarrow
AH=AD\cos 60^o=\frac{1}{2}|\overrightarrow{b}|$
Do hình
thang ABCD cân $\Rightarrow AM=CD=AB-2AH=|\overrightarrow{a}|-|\overrightarrow{b}|$
$\Rightarrow
BM=AB-AM=|\overrightarrow{b}|$
$\mathop {BC}\limits^ \to =\mathop {BM}\limits^ \to+\mathop {MC}\limits^ \to=-\frac{\mathop {|b|}\limits^ \to }{\mathop {|a|}\limits^ \to } \mathop {a}\limits^ \to +\mathop {b}\limits^ \to $
$\mathop {AC}\limits^ \to =\mathop {AD}\limits^ \to +\mathop {AM}\limits^ \to =\mathop {b}\limits^ \to +\mathop {a}\limits^ \to -\frac{\mathop {|b|}\limits^ \to }{\mathop {|a|}\limits^ \to } \mathop {a}\limits^ \to $
$\mathop {BD}\limits^ \to =\mathop {BA}\limits^ \to +\mathop {AD}\limits^ \to =\mathop {b}\limits^ \to -\mathop {a}\limits^ \to $
$\mathop {AC}\limits^ \to \bot \mathop {BD}\limits^ \to $$\Leftrightarrow \mathop {AC}\limits^ \to .\mathop {BD}\limits^ \to =0 \Leftrightarrow(\mathop {b}\limits^ \to +\mathop {a}\limits^ \to -\frac{\mathop {|b|}\limits^ \to }{\mathop {|a|}\limits^ \to }\overrightarrow{a} ) (\mathop {b}\limits^ \to -\mathop {a}\limits^ \to ) =0$
$\Leftrightarrow 2a^2-2ab-b^2=0$(do $\overrightarrow{a}.\overrightarrow{b}=a.b.\cos60^o$)
$\Rightarrow a=\frac{b}{2} (1+\sqrt{3} )$