Let ABC be a circle, of which D is the centre and AB a diameter: the straight line drawn at right angles to AB, from its extremity A, shall fall without the circle.
For, if not, let it fall, if possible, within the circle, as AC, and draw DC to the point C, where it meets the circmference.
Then, because DA is equal to DC, [I. Definition 15.
the angle DAC is equal to the angle DCA. [I. 5.
But the angle DAC is a right angle; [Hypothesis.
therefore the angle DCA is a right angle;
and therefore the angles DAC, DCA are equal to two right angles; which is impossible. [I. 17.
Therefore the straight line drawn from A at right angles to AB does not fall within the circle.
And in the same manner it may be shewn that it does not fall on the circumference.
Therefore it must fall without the circle, as AE.
Also between the straight line AE and the circumference, no straight line can be drawn from the point A, which does not cut the circle.
For, if possible, let AF be between them; and from the centre D draw DG perpendicular to AF; [I. 12.
let DG meet the circumference at H.
Then, because the angle DGA is a right angle, [Construction.
the angle DAG is less than a right angle; [I. 17.
therefore DA is greater than DG. [1.19.
But DA is equal to DH; [I. Definition 15.
therefore DH is greater than DG, the less than the greater;
which is impossible.
Therefore no straight line can be drawn from the point A between AE and the circumference, so as not to cut the circle.
