Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the circle.
Proof
Let ABC be the given circle and D the given segment, not greater
than the diameter BC of the circle. Apply I.3 to cut off from BC
a segment BE equal to D. With centre B and radius BE describe
a circle EAF (Postulate 3); join BA where EAF meets the given
circle. Then BA=BE=D by Definition I.15, so BA is the required
chord.