Question

求各位大神帮帮忙看看，要过程，谢谢

Answer 1

设AC和DD'交于点E
∵∠BAC=∠BAD+∠CAD,∠DAD'=∠CAD'+∠CAD
∵∠BAD=∠CAD'
∴∠BAC=∠DAD'
∵AD=AD'
∴∠ADD'=∠AD'D=1/2(180-∠DAD')
∵∠ABC=∠ECP=1/2(180-∠BAC),∠ADE=∠AD'E=1/2(180-∠DAD'),∠BAC=∠DAD'
∴∠ADE=∠AD'E=∠ABC=∠ECP
∵∠PEC=∠AED',∠AD'E=∠ECP
∴△AD'E∽△PCE
∴AE/PE=D'E/CE, ∠CAD'=∠CPE
∵∠AEP=∠CED',AE/PE=D'E/CE
∴△APE∽△CED'
∴∠ACD'=∠APE
∵∠ACD'+∠ACD'=90 ,∠CAD'=∠CPE ,∠ACD'=∠APE
∴∠CPE +∠APE=90
∵∠APC=∠CPE +∠APE
∴∠APC=90
∵AB=AC,∠APC=90,由等腰三角形三线合一得
∴BP=CP

Answer 2

连接AP

∵ AD⊥BD

∴ ∠2+∠5=90°

∵AD=AD′ AB=AC  ∠DAD′=∠BAC

∴∠5=∠ABC

∠ABC+∠BAC/2=90°

∴∠2=∠BAC/2

∠ABC-∠1+∠4=90° ∠1=∠ABC+∠4-90°

∠3=∠1+∠2=∠ABC+∠4-90°+∠BAC/2=∠4

∴APC D′四点共圆

∵AD′⊥CD′

∴AP⊥PC

∴BP=CP

Answer 3

写的有点乱，将就着看吧

Answer 4

诸位大神牛逼啊