⊙O是△ABC的外接圆,圆心O在AB上,过点B作⊙O的切线交AC的延长线于点D.求证△ABC∽△BDC
4个回答

(1)

根据已知条件得

△ABC为RT△,∠C=90

RT△ABC与RT△ABD共用∠A

RT△ABC∽RT△ABD

同理可求

RT△BDC∽RT△ABD

RT△BDC∽RT△ABC

(2)

AC=8,BC=6

根据勾股定理得,AB=10

RT△ABC∽RT△ABD

AC:BD=BC:AB

8:BD=6:10

BD=40/3

S△ABD=1/2ABxBD=1/2x10x40/3=200/3

S△ABC=1/2ACxBC=1/2x8x6=24

S△BCD=S△ABD-△ABC=200/3-24=128/3