如图,在梯形ABCD中,AD∥BC,∠ABC=90°,DG⊥BC于G,BH⊥DC于H,CH=DH,点E在
![](http://hiphotos.baidu.com/zhidao/pic/item/37d3d539b6003af3a94d2fd1362ac65c1138b688.jpg)
AB上,点F在BC上,并且EF∥DC.
(1)若AD=3,CG=2,求CD;
(2)若CF=AD+BF,求证:EF=
CD.
(1)连BD,如图,∵在梯形ABCD中,AD∥BC,∠ABC=90°,DG⊥BC,∴四边形ABGD为矩形,∴AD=BG=3,AB=DG,又∵BH⊥DC,CH=DH,∴△BDC为等腰三角形,∴BD=BG+GC=3+2=5,在Rt△ABD中,AB=BD2−AD2=52−32=4,∴DG=4,...