如图,正方形为ABCD,折线为EF.由题意知EF=AB.作AG‖EF交DC于G,则AEFG为平行四边形,AG=EF,得AG=AB.显然,△ADG∽△BAD.则AG/AD=BD/AB.设AD=a, AB=AG=b, BD=c.代入上式有：b/a=c/b将上式两边同乘以b/a得：b{%&...