问题补充:
在△ABC中,点D、E分别在AB、AC边上,∠ADE=∠C,BD=AE+AD,CE=AE-AD,求BC:DE的值
答案:
∠ADE=∠C;∠A=∠A;==>△ABC∽△AED
==>AC/AD=AB/AE=BC/DE;
==>(AC+AB)/(AD+AE)=(AB-AC)/(AE-AD)=BC/DE;
∵BD=AE+AD,CE=AE-AD;
∴AC+AB/BD=AB-AC/CE=BC/DE;
==>(AD+BD+AE+CE)/BD=(AD+BD-AE-EC)/CE=BC/DE;
==>(2BD+CE)/BD=(BD-2CE)/CE=BC/DE;
==>BD^2-4BD*CE-CE^2=0;
==>BD=(2+根号5)CE;
==>BC/DE=(BD-2CE)/CE=[(2+根号5)CE-2CE]/CE=根号5;
BC:DE的值为根号5;
在△ABC中,点D、E分别在AB、AC边上,∠ADE=∠C,BD=AE+AD,CE=AE-AD,求BC:DE的值(图1)答案网 答案网