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# In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show thati) $$\triangle{APD}$$ $$\displaystyle \cong$$ $$\triangle{CQB}$$ ii) AP = CQiii) $$\triangle{AQB}$$ $$\displaystyle \cong$$ $$\triangle{CPD}$$ iv) AQ = CPv) 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)
(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.