In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that
i) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\)
ii) AP = CQ
iii) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\)
iv) AQ = CP
v) APCQ is a parallelogram.

Given, ABCD is a parallelogram and P and Q are lie on BD such that DP = BQ ...(i)

i) Now,in \(\triangle{APD}\) and \(\triangle{CQB}\), we have,
DP = BQ ...(Given)
AD = BC
(Opposite sides are equal in parallelogram)
\(\angle{ADP}\) = \(\angle{QBC}\)
(Since, AD || BC and BD is a transversal)
\(\therefore\) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\) ...(By SAS rule)

ii)
\(\because \) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\)
\(\therefore \) AP = CQ ...(By CPCT)

iii) Here, Now, in \(\triangle{AQB}\) and \(\triangle{CPD}\), we have,
DP = BQ ...(Given)
AB = CD
(Opposite sides are equal in parallelogram)
\(\angle{ABQ}\) = \(\angle{CDP}\)
(Since, AB || CD and BD is a transversal)
\(\therefore \) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\) ...(By SAS rule)

iv)
\(\because \) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\)
\(\therefore \) AQ = CP...(By CPCT)

v) Now,in \(\triangle{APQ}\) and \(\triangle{PCQ}\), we have,
AQ = CP (From part iv))
AP = CQ (From part ii))
PQ = QP ...(Common side)
\(\therefore \) \(\triangle{APQ}\) \(\displaystyle \cong \) \(\triangle{PCQ}\) ...(By SSS rule)
\(\therefore \) \(\angle{APQ}\) = \(\angle{PQC}\)
And \(\angle{AQP}\) = \(\angle{CPQ}\)
(Vertically Opposite angles)

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.

\(\therefore \) AP || CQ and AQ || CP

Now, both pairs of opposite sides of quadrilateral APCQ are parallel.

Hence, it is proved that APCQ is a parallelogram.