令 |BD| = t;则有
|BC| = √3 t; |AB| = √[t² - 1]; |DC| = √3 t - t
△ABC中,余弦定理:cos(B) = (|AB|² + |BC|² - |AC|²)/(2|AB||AC|)
△ABD中,cos(B) = |AB|/|BD|;
综合上式,解得 |AC|² = 4t² - 2√3 t² + 2√3 - 1;
△ADC中,余弦定理:cos(∠DAC) =(|AD|² + |AC|² - |DC|²)/(2|AD||AC|)
向量AC*向量AD = |AD||AC|cos(∠DAC) = (|AD|² + |AC|² - |DC|²)/2;
将|AC|²代入上式可以求出:向量AC*向量AD = 2√3