已知 $\triangle ABC$ 满足 $AB=1$，$AC=2$，$\cos A=\dfrac 7{25}$．若 $E$ 为 $\triangle ABC$ 内一点，满足 $\lambda\overrightarrow{AE}=2\overrightarrow{AB}+\overrightarrow{AC}$（$\lambda\in\mathbb R$），且 $\overrightarrow {EB}\cdot \overrightarrow{EC}=0$，延长 $AE$ 至 $BC$ 交于点 $D$，则 $\dfrac{AD}{\lambda}=$ _______．
答案 $\dfrac{6-\sqrt{22}}{15}$．
解析 统一起点为 $A$，则\[\overrightarrow {EB}\cdot \overrightarrow{EC}=0\iff \left(\overrightarrow {AB}-\overrightarrow {AE}\right)\cdot \left(\overrightarrow {AC}-\overrightarrow{AE}\right)=0,\]记 $\dfrac 1{\lambda}=x$，则\[\overrightarrow{AE}=2x\overrightarrow{AB}+x\overrightarrow{AC},\]于是\[\left((1-2x)\overrightarrow{AB}-x\overrightarrow{AC}\right)\cdot \left(-2x\overrightarrow{AB}+(1-x)\overrightarrow{AC}\right)=0,\]从而\[-2x(1-2x)\cdot 1^2-x(1-x)\cdot 2^2+(2x^2+(1-2x)(1-x))\cdot 1\cdot 2\cdot \dfrac7{25}=0,\]即\[128x^2-64x+7=0,\]解得 $x=\dfrac{6-\sqrt{22}}{16}$．而\[AD=\sqrt{\left(\dfrac 23\overrightarrow{AB}+\dfrac 13\overrightarrow{AC}\right)^2}=\sqrt{\dfrac 49\cdot 1^2+\dfrac 19\cdot 2^2+\dfrac 49\cdot 1\cdot 2\cdot \dfrac7{25}}=\dfrac{16}{15},\]从而\[\dfrac{AD}{\lambda}=AD\cdot x=\dfrac{6-\sqrt{22}}{15}.\]