$\dfrac{AH}{I'E}=\dfrac{AH}{\dfrac{a·AH}{b+c-a}}=\dfrac{b+c-a}a$
$\dfrac{AI}{I'D}=\dfrac{2\dfrac{AI}{AI'}}{1-\dfrac{AI}{AI'}}=\dfrac{2\dfrac{b+c-a}{a+b+c}}{1-\dfrac{b+c-a}{a+b+c}}=\dfrac{b+c-a}a$
$\therefore\frac{AH}{I'E}=\frac{AI}{I'D}$.
又AH∥I'E,∴△AIH~△I'DE,∴DE∥HI |