If AD and BC are extended to meet at R we get a triangle ARB. The triangles ARB, PRQ and DRC are similar because AB, PQ and DC are parallel. RD/RP=**DC/PQ**=RC/RQ; RD/RA=DC/AB=RC/RB; RP/RA=**PQ/AB**=RQ/RB. We have two equations containing PQ (in bold print). However, we don't have any more information about the position of PQ in relation to AB or CD. All we know is DC<PQ<AB.

The middle equation merely proves the given theorem that if RD/RA=1/2 (D is the midpoint of RA) then DC/AB=1/2, DC=(1/2)AB.

If P is the midpoint of AD then DP=PA. Let RA=4x. Then RD=2x, RP=3x, RD/RP=2/3=DC/PQ; RP/RA=3/4=PQ/AB. Therefore PQ=3AB/4=3DC/2 and DC=AB/2.