Question

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=

Answer 1

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