Prove that $DQ \times DB = DP \times DC + DR \times DA$.

Question

Let $ABCD$ be a parallelogram, with $P$, $Q$, and $R$ the points on which a given circle passes through $D$ and cuts through the segments $CD$, $BD$ and $AD$ respectively:

How do you prove that $DQ \times DB = DP \times DC + DR \times DA$?

Answer 1

Since $\angle PRQ = \angle PDQ = \angle ABD$ and $\angle RQP = 180^\circ - \angle PDR = \angle DAB$, triangles $PQR$ and $DAB$ are similar. Therefore there exists $k>0$ such that $$k = \frac{PQ}{DA} = \frac{QR}{AB} = \frac{RP}{BD}.$$ By Ptolemy theorem in $DRQP$ we get $$DQ \cdot PR = DP \cdot RQ + DR \cdot PQ.$$ Substituting $PR = k\cdot DB, \ QR = k \cdot AB = k \cdot DC, \ PQ = k \cdot DA$ and dividing by $k$ we arrive at $$DQ \cdot DB = DP \cdot DC + DR \cdot DA.$$