如图,在Rt△ABC中,∠B=90°,它的内切圆分别与边BC、CA、AB相切于点D、E 、F,连接AD与内切圆相交于另

2024-11-17 19:30:11
推荐回答(1个)
回答1:

(1)∵BC与圆相切,
∴∠PFD=∠PDC.


∵BF、BD分别于圆相切,
∴∠BFD=∠BDF=45°.
∴∠FPD=45°.
∵PC⊥PF,
∴∠FPD=∠DPC.
∴△PFD △PDC.

(2)∵AE、AF与圆相切,
∴∠AFP=∠ADF,∠AEP=∠ADE,
∵∠FAD=∠PAF,∠EAP=∠DAE,
∴△AFP △ADF,△AEP △ADE,
AF
AD
=
PF
FD
AE
AD
=
PE
ED
且AE=AF,
PF
FD
=
PE
ED

∵△PFD △PDC,
PF
FD
=
PD
DC

EP
DE
=
PD
DC