Final answer:
Through the properties of parallelograms and similar triangles, we can prove that triangles ∆APD and ∆COB are similar, leading to the equalities AP = CQ, ∠AQB = ∠CPD, and AQ = CP. These equalities confirm that APCQ itself is a parallelogram.
Step-by-step explanation:
To show the properties of points P and Q on the diagonal BD of parallelogram ABCD where DP = BQ, we need to consider the properties of parallelograms and their diagonals, along with the properties of similar triangles.
(i) To show that ∆APD ~ ∆COB, we observe that ∆APD and ∆COB are both right triangles that share the angle at vertex C/D since ABCD is a parallelogram and opposite angles are equal. Moreover, since DP = BQ and AD = BC (parallelogram sides), the triangles are similar by the Angle-Angle-Side (AAS) criterion.
(ii) For AP = CQ, we use the similarity of the triangles once more to understand the proportionality of sides. As the triangles are similar, the corresponding sides are proportional, which means AP/AD = CB/CQ. Since AD = BC in a parallelogram, this implies AP = CQ.
(iii) ∠AQB = ∠CPD because in similar triangles, corresponding angles are equal. This stems from the similarity of ∆APD and ∆COB proven in (i).
(iv) To show AQ = CP, we look at ∆AQB and ∆CPD, which are also similar. By the Side-Angle-Side (SAS) criterion, the shared angles and the known proportional sides (AB = CD and AP = CQ) give us that AQ/AB = CP/CD, and thus AQ = CP.
(v) APCQ forms a parallelogram as opposite sides are equal (AP = CQ and AQ = CP) and opposite angles are equal (∠AQB = ∠CPD), which are basic properties for parallelograms.
Throughout this proof, we utilize the concepts of similar triangles, properties of parallelograms, and the Angle-Angle-Side criterion to justify the similarity of triangles and the equality of sides and angles necessary to conclude APCQ is a parallelogram.