三角形ABC中,AD=BC,AD垂直BC,BE垂直AC,G为BC中点.求证：三角形BDF相似于三角形ADC.DF+GF=

综合应用相似和勾股定理
证明：
△BDF∽△ADC
DF/BD=DC/AD
DF=BD*DC/AD
BD=BG+GD ,DC=GC- GD=BG - GD ,AD=BC ,
勾股定理
FG^2
=FD^2+DG^2
=(BD*DC/AD)^2+DG^2
BD=BG+GD ,DC=GC- GD=BG - GD ,AD=BC ,
=[(BG +GD)(BG - GD) / BC]^2+DG^2
=(BG^2- GD^2)^2 / BC^2+BC^2*DG^2 / BC^2
=[(BG^2- GD^2)^2 +BC^2*DG^2] / BC^2
=[(BG^2- GD^2)^2 +(2BG)^2*DG^2] / BC^2
=[(BG^2- GD^2)^2 +4 BG^2*DG^2] / BC^2
=(BG^2+GD^2)^2 / BC^2
FG=(BG^2+GD^2 ) / BC
FG+DF
=(BG^2+GD^2 ) / BC+BD*DC/AD
=(BG^2+GD^2 ) / BC+[(BG +GD)(BG - GD)/BC
=(BG^2+GD^2 ) / BC+(BG^2 - GD^2)/BC
=2BG^2 / BC
=2BG^2 /2BG
=BG
=1/2 BC

• AD垂直BC,BE垂直AC，角C+角EBC=90°，角C+角CAD=90°，所以角EBC=角CAD，又因为角BDF=角ADC，所以三角形BDF相似于三角形ADC