therefore KB,KG, KF all equal to each other. [I. Def. 15.
And because within the circle DEF, the point K is taken, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF, therefore K is the centre of the circle DEF. [III. 9.
But K is also the centre of the circle ABC. [Construction.
Therefore the same point is the centre of two circles which cut one another;
which is impossible. [III. 5.
Wherefore, one circumference &c. q.e.d.
PROPOSITION 11. THEOREM.
If two circles touch one another internally, the straight line which joins their centres, being produced, shall pass through the point of contact.
Let the two circles ABC, ADE touch one another internally at the point A; and let F be the centre of the circle ABC, and G the centre of the circle ADE: the straight line which joins the centres F, G, being produced, shall pass through the point A .
For, if not, let it pass otherwise, if possible, as FGDH, and join AF,AG.
Then, because AG, GF are greater than AF, [I. 20.
and AF is equal to HF, [I. Def. 15. therefore AG, GF, are greater than HF. Take away the common part GF; therefore the remainder AG is greater than the remainder HG. But AG os equal to DG. [I. Definition 15.
Therefore DG is, greater than HG, the less than the greater;
which is impossible.
Therefore the straight line which joins the points F, G, being produced, cannot pass otherwise than through the point A, that is, it must pass through A.
Wherefore, if two circles &c. q.e.d.
