P and Q are the points of trisection of the diagonal AC of a parallelogram ABCD.
Prove that BQ || DP and BD bisects PQ.

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:

(i)    ∆APD ≅ ∆CQB
(ii)   AP = CQ
(iii)  ∆AQB ≅ ∆CPD
(iv)  AQ = CP
(v)   APCQ is a parallelogram.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (see figure). Show that:
(i) ∆APB ≅ ∆CQD
(ii) AP = CQ.

In ∆ABC and ∆DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, Band C are joined to vertices D, E and F respectively (see figure). Show that:
(i)     quadrilateral ABED is a parallelogram
(ii)    quadrilateral BEFC is a parallelogram
(iii)   AD || CF and AD = CF
(iv)   quadrilateral ACFD is a parallelogram

(v)     AC = DF
(vi)    ∆ABC ≅ ∆DEF. [CBSE 2012

ABCD is a trapezium in which AB || CD and AD = BC (see figure): Show that
(i)      ∠A = ∠B
(ii)    ∠C = ∠D
(iii)    ∆ABC = ∆BAD
(iv)    diagonal AC = diagonal BD.



[Hint. Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]