¶ An other demonstration after Pelitarius and Flussates.
Suppose that there be two circles ABG and ADG, which if it be possible, let touch the one the other outwardly in moe poyntes then one, namely, in A and G. Let the centre of the circle ABG be the poynt I, and let the centre of the circle ADG be the poynt K. And draw a right line from the poynt I to the poynt K, which (by the 12. of thys booke) shall passe both by the
poynt A and by the poynt G: which is not possible: for then two right lines should include a superficies, contrary to the last common sentence. It may also be thus demonstrated. Draw a line from the centre I to the centre K, which shall passe by one of the touches, as for example by the poynt A. And draw these right lines GK and GI, and so shall be made a tri∣angle, whose two sides GK and GI shall not be greater then the side IK: which i
•• contrary to the 20. of the first.
But now if it be possible, let the foresayd circle ADG touch the circle ABC inward∣ly in moe poyntes then one, namely, in the pointes A and G: and let the centre of the circle ABC be the poynt I, as before: and let
the centre of the circle ADG be the poynt K, as also before. And extend a line from the poynt I to the poynt K, which shall fall vpon the touch (by the 11. of thys booke). Draw also these lines KG, and IG. And for asmuch as the line KG is equall to the line KA (by the 15. definition of the first) adde the line KI common to them both. Wherefore the whole line AI is equall to the two lines KG and KI: but vnto the line AI is equall the line IG (by the definition of a circle). Wher∣fore in the triangle IKG the side IG is not lesse then the two sides IK and KG: which is con∣trary to the 20. of the first.