如图,已知∠A=∠B,AA1,PP1,BB1均垂直于A1B1,AA1=17,PP1=16,BB1=20,A1B1=12,则AP+PB等于( ) A.12 B.13 C.14 D.15
问题描述:
如图,已知∠A=∠B,AA1,PP1,BB1均垂直于A1B1,AA1=17,PP1=16,BB1=20,A1B1=12,则AP+PB等于( )
A. 12
B. 13
C. 14
D. 15
答
如图,AA1,PP1,BB1均垂直于A1B1,
∴AA1∥PP1∥BB1,
过点P作PF⊥AA1,交AA1于点D,交BB1于点F,延长BP交AA1于点C,作CG⊥BB1,交BB1于点G,
∴四边形DFB1A1,DPP1A1,FPP1B1,FDGC,CGB1A1是矩形,
∴DA1=PP1=FB1=16,CG=A1B1=12,
∵AA1∥BB1,
∴∠B=∠ACB,
∵∠A=∠B
∴∠A=∠BCA,
∴AP=CP,
∵PF⊥AA1,
∴点D是AC的中点,
∵AA1=17,
∴AD=CD=17-16=1,BF=20-16=4,FG=CD=1,BG=4+1=5,
∴BP+PA=BP+PC=BC=
=
CG2+BG2
=13.
122+52
故选B.