如图,直角三角形ABC,角平分线AD,BC=2AC,求证ab+2bd=5ac

问题描述:

如图,直角三角形ABC,角平分线AD,BC=2AC,求证ab+2bd=5ac

做DF垂直于AB于F点
AB+2BD
=AF+FB+2BD
=AC+2BD+FB
因为 BFD与BCA相似,所以 2FD=FB
所以
AC+2BD+FB
=AC+2BD+2FD
=AC+2CB
=AC+4AC
=5AC
请采纳答案,支持我一下.