from D let DF be drawn at right angles to EA, and let EF, AB be joined; I say that AB has been drawn from the point A touching the circle BCD.
For, since E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two sides FE, ED: and they contain a common angle, the angle at E; therefore the base DF is equal to the base AB, and the triangle DEF is equal to the triangle BEA, and the remaining angles to the remaining angles; [[I. 4](elem.1.4)] therefore the angle EDF is equal to the angle EBA.

But the angle EDF is right; therefore the angle EBA is also right.

Now EB is a radius; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [[III. 16, Por.](elem.3.16.p.1)] therefore AB touches the circle BCD.

Therefore from the given point A the straight line AB has been drawn touching the circle BCD.