If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle.
Proof
Let A, B be on the circle with centre E (III.1). Suppose for
contradiction that some point F on the chord AB lies outside the
circle; then EF>EA. Join EA, EB. By I.5, the base angles
of the isoceles △EAB are equal: ∠EAB=∠EBA. By I.16, the exterior angle at any interior point F of
AB is greater than either remote interior angle; pursuing the
inequalities (Heath's argument) forces EF<EA for F inside
AB, contradicting the assumption. Hence every point of AB lies
within the circle.