Question

已知：Rt三角形abc中,角a=30度,d在直角边ab上,bd：ad=1：2.求证：dc平分角bca

Answer 1

AB^2+BC^2=AC^2-------------------------^2为平方因为AC=2BC所以AB^2+BC^2=4BC^2得3BC^2=AB^2又AB=3BD所以3BC^2=9BD^2 得BC^2=3BD^2又BD^2+BC^2=CD^2所以 BD^2+3BD^2=CD^2得 4BD^2=CD^2所以 CD=2BD=AD所以 角A=角DC...