
(1)證明:連接PB,OP,
∵PE⊥AB,PD⊥OB,
∴∠BEP=∠PDO=90°,
∵AB切⊙O
1于B,∠ABP=∠BOP,
∴△PBE∽△POD,
∴

=

,
同理,△OPF∽△BPD
∴

=

,
∴

=

,
∴PD
2=PE•PF;
(2)解:連接O
1B,O
1P,
∵AB切⊙O
1于B,∠POB=30°,
∴∠ABP=30°,
∴∠O
1BP=90°-30°=60°,
∵O
1B=O
1P,
∴△O
1BP為等邊三角形,
∴O
1B=BP,
∵P為弧BO的中點,
∴BP=OP,
即△O
1PO為等邊三角形,
∴O
1P=OP=a,
∴∠O
1OP=60°,
又∵P為弧BO的中點,
∴O
1P⊥OB,
在△O
1DO中,∵∠O
1OP=60°O
1O=a,
∴O
1D=

a,OD=

a,
過D作DM⊥OO
1于M,∴DM=

OD=

a,
OM=

DM=

a,
∴D(-

a,

a),
∵∠O
1OF=90°,∠O
1OP=60°
∴∠POF=30°,
∵PE⊥OA,
∴PF=

OP=

a,OF=

a,
∴P(-

a,

),F(xiàn)(-

a,0),
∵AB切⊙O
1于B,∠POB=30°,
∴∠ABP=∠BOP=30°,
∵PE⊥AB,PB=a,
∴∠EPB=60°
∴PE=

a,BE=

a,
∵P為弧BO的中點,
∴BP=PO,
∴∠PBO=∠BOP=30°,
∴∠BPO=120°,
∴∠BPE+∠BPO=120°+60°=180°,
即OPE三點共線,
∵OE=

a+a=

a,
過E作EM⊥x軸于M,∵AO切⊙O
1于O,
∴∠EOA=30°,
∴EM=

OE=

a,OM=

a,
∴E(-

a,

a),
∵E(-

a,

a),D(-

a,

a),
∴DE=-

a-(-

a)=

a,
DE邊上的高為:

a,
∴S
△DEF=

×

a×

a=

a
2.
故答案為:D(-

a,

a),E(-

a,

a),F(xiàn)(-

a,0),P(-

a,

);S
△DEF=

a
2.
分析:(1)連接PB,OP,利用AB切⊙O
1于B求證△PBE∽△POD,得出

=

,同理,△OPF∽△BPD,得出

=

,然后利用等量代換即可.
(2)連接O
1B,O
1P,得出△O
1BP和△O
1PO為等邊三角形,根據(jù)直角三角形的性質(zhì)即可解得D、E、F、P四個點的坐標(biāo).
再利用三角形的面積公式可直接求出三角形DEF的面積.
點評:本題主要考查學(xué)生對相似三角形的判定與性質(zhì),切割線定理,坐標(biāo)與圖形性質(zhì)等知識點的理解和掌握,此題綜合性強,難度較大,屬于難題.