(Taiyuan Competition in 2002) As shown in the figure, the position χ o of AB is known, and the position χ o of C is slightly higher, so that BC extends to D, and CD=BC, CE⊥AD is in E, E and BE is in O

(Taiyuan Competition in 2002) As shown in the figure, the position χ o of AB is known, and the position χ o of C is slightly higher, so that BC extends to D, and CD=BC, CE⊥AD is in E, E and BE is in O F. When OC is connected, OC is the center line of △DAB, so OC‖AD and CE⊥AD, so EC⊥CO, so we know that PC is tangent to circle O, so pc 2 = pf× pa.

And PE⊥AE, EF⊥PA, PE 2 = PF× PA are known from the projective theorem.

So PE=PC.