解答:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9e3df8dcd100baa1828f003c4410b912c9fc2e91?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
(1)解:过点P作PQ⊥AB于点Q.
∵PA=PB,∠APB=120°,AB=4
∴AQ=BQ=2
,∠APQ=60°(等腰三角形的“三线合一”的性质),
在Rt△APQ中,sin∠APQ=
∴AP=
=
=
=4;
(2)证明:过点P分别作PS⊥OM于点S,PT⊥ON于点T.
∴∠OSP=∠OTP=90°(垂直的定义);
在四边形OSPT中,∠SPT=360°-∠OSP-∠SOB-∠OTP=360°-90°-60°-90°=120°,
∴∠APB=∠SPT=120°,∴∠APS=∠BPT;
又∵∠ASP=∠BTP=90°,AP=BP,
∴△APS≌△BPT,
∴PS=PT(全等三角形的对应边相等)
∴点P在∠MON的平分线上;
(3)①∵OP平分∠AOB,∠AOB=60°,OP⊥AB,
∴AQ=BQ=
AB=2
,
∴
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/0ff41bd5ad6eddc43344f37c3adbb6fd53663391?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
OQ=
=6,
同理:PQ=
=2,
∴OP=8,
∵点C,D,E,F分别是四边形AOBP的边AO,OB,BP,PA的中点,
∴CD=EF=
AB,CF=DE=
OP,
∴四边形CDEF的周长为:8+4
②CD和EF是△ABO和△ABP的中位线,
则CD=EF=
AB=2
,
CF和DE分别是△AOP和△BOP的中位线,则CF=DE=
OP,
当AB⊥OP时,OP为四点边形AOBP外接圆的直径时,OP最大,其值是8,OP一定大于当点A或B与点O重合时的长度是4.
则4+4
<t≤8+4
.