Then, since AG, GF are greater than FA, that is, than FH,
let FG be subtracted from each;
therefore the remainder AG is greater than the remainder GH.
But AG is equal to GD;
therefore GD is also greater than GH, the less than the greater: which is impossible.
Therefore the straight line joined from F to G will not fall outside;
therefore it will fall at A on the point of contact.