colourxb
幼苗
共回答了22个问题采纳率:90.9% 举报
∠C=90°,AC=3,AB=5
BC=4
ΔACD≌ΔAC1D
AC=AC1=3,设CD=C1D=t
BC1=5-3=2 BD=4-t
BD^2=BC1^2+C1D^2
(4-t)^2=4+t^2==>t=3/2
AD^2=3^2+(3/2)^2=45/4
AD=3√5/2
若APD是等腰三角形
则 AP=AD=3√5/2
或AP=PD 或AD=PD
当AP=PD 时,
PD^2-PC1^2=DC1^2=9/4
AP^2-(3-AP)^2=9/4
==>AP=15/8
当AD=PD 时,AC1=PC1=3,AP=6
∴AP=15/8,或3√5/2,或AP=6
1年前
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