Question

如图，过平行四边形ABCD的顶点A的直线交BD于点P，交CD于点Q，并交BC的延长线于点R。求证：AP²=PQ×PR

Answer 1

过P作AB的平行线
交BC于E
设BE=X，EC=Y，CR=Z
因为△RPE∽△RAB，△BPE∽△BDC
PE/AB=(Y+Z)/(X+Y+Z)，PE/DC=X/(X+Y)
又AB=DC
所以(Y+Z)/(X+Y+Z)=X/(X+Y)
得X²=Y(Y+Z)
即Y/(Y+Z)=Y²/X²
Y/X=PD/PB，Y/(Y+Z)=PQ/PR
PQ:QR=PD²:PB²

Answer 2

AB/DQ = AP / PQ = BP / PD; PR /AP = BR / AD = BP /PD;所以 AP / PQ = PR /AP

Answer 3

∵AD∥BR
∴△APD∽△BPR
∴AP:PR=DP:PB
又AB∥QD
∴△ABP∽△DPQ
∴PQ:AP=DP:PB
∴AP:PR=PQ:AP
∴AP²=PQ×PR

Answer 4

看数学书