Problem: In this figure the center of the circle is O. ABβ₯BC, ADOE is a straight line, AP=AD, and AB has a length twice the radius. Then:
Answer Choices:
A. AP2=PBβ
AB
B. APβ
DO=PBβ
AD
C. AB2=ADβ
DE
D. ABβ
AD=OBβ
AO
E. none of these
Solution:
Since ABADβ=AEABβ,AB2=ADβ
AE
But AE=AD+2r=AD+ABβ΄AB2=AD(AD+AB)
=AD2+ADβ
AB
β΄AD2=AB2βADβ
AB=AB(ABβAD)
Since AP=AD,AP2=AB(ABβAP)=ABβ
PB