几何证明题,要经过,谢谢

1个回答

  • (1)RT△ABC中,∠C=90,∠B=30.所以∠CAB=60

    AD平分∠CAB,所以∠BAD=∠CAB/2=30

    ∠BAD=∠B,所以AD=DB

    (2)△ABC和△AEF中

    ∠A=∠A,∠AFE=∠C=90

    所以△ABC∽△AEF,AE:AB=AF:AC

    RT△ABC中,∠B=30,所以AB=2AC=12

    AE=AC-CE=6-X,AF=AB-BF=12-Y

    (6-X):12=(12-Y):6

    6-X=2(12-Y)

    6-X=24-2Y

    2Y=18+X

    Y=(18+X)/2=X/2+9

    (3)RT△ACD中,∠CAD=∠CAB/2=30

    所以AC:CD=√3,CD=2√3

    由(2)中三角形相似得:∠AEF=∠B=30

    RT△AEF中,AE=2AF

    ∠DEF=90,所以∠DEC=180-90-30=60

    RT△CDE中,CD=√3CE,所以CE=2

    AE=AC-CE=4

    AF=AE/2=2

    BF=AB-AF=10