Problem. In a given circle (ABC) to place a chord equal to a given line (D) not greater than the diameter.
Sol.—Draw any diameter AC of the circle; then, if AC be equal to D, the thing required is done; if not, from AC cut off the part AE equal to D [I. iii.]; and with A as centre and AE as radius, describe the circle EBF, cutting the circle ABC in the points B, F. Join AB. Then AB is the chord required.
Dem.—Because A is the centre of the circle EBF, AB is equal to AE [I. Def. xxxii.]; but AE is equal to D (const.); therefore AB is equal to D.