解:(1) ∵平行四边形ABCD中,AD//BC ∴∠DAB+∠ABC=180° ∵AP 与BP分别平分∠DAB和∠CBA ∴∠PAB=0.5∠DAB ∠PBA=0.5∠ABC ∴∠PAB+∠ABP=90°△APB是直角三角形 (2)∵DC//AB AD=BC ∴∠APD=∠PAB=∠PAD ∴AD=PD 同理BC=PC ∴PD=PC (3)∵∠AEB=∠APC=90°∠EAF=∠PAB ∴ △AEF∽△APB ∵ AB=DC=2AD=100mm AP=80mm ∴ BP=60mm ∴ tan∠ABP=tan∠AFE= 4/3 (4)作过切点G半径O"G,O"G//AE,O"M//PB O"G/AE+O"M/BP=O"B/AB+O"A/AB=1 设EF=3m 则AM=5m,AP=80-5m tan∠BFP=tan∠AFE= 4/3 得60/(80-5m)= 4/3 m=7 AE= 4m=28 r/28+r/60=1 解得r=210/11 | |