Question

在梯形ABCD中,AD平行于DC,AD大于BC,且AD=5,AB=AC=2,P为AD上一点,且角BPC=角A,求AP=?

Answer 1

∵∠BPD=∠BPC+∠CPD=∠A+∠ABP,∠BPC=∠A

∴∠CPD=∠ABP

∵∠D=∠A

∴△ABP∽△DPC

∴AP/CD=AB/PD

∴AB*CD=AP*PD

设AP=x

则4=x(5-x)

x²-5x+4=0

x1=1,x2=4

所以AP=1或AP=4