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
1个回答

(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 .