| http://www.w3.org/ns/prov#value | - . (Draw a figure to convince your self.) Here, however, is a simple proof of the original problem: Consider first the circles BPC, CPA. If the third circle APB has common points with the lens like common part of the two above circles, then some point of the circumference of circle APB will cut arc PC somewhere in between P and C. Call this pointQ.Note then that circles APB and BPC contain Q. In ot
|
