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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.

Answer :

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.

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