P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP.Prove also that AC bisects PQ.
Answers
AP ║ CQ & AC bisects PQ
Step-by-step explanation:
P and Q are the points of trisection of the diagonal BD of a parallelogram
=> BP = PQ = QD = BD/3
BQ = BP + PQ = 2BD/3
PD = PQ + QD = 2BD/3
in ΔAPD & ΔBQC
AD = BC ( opposite sides of Parallelogram)
PD = QC ( shoen above)
∠ADP = ∠CBQ ( as ∠ADB = ∠CBD and ∠ADP = ∠ADB & ∠CBQ = ∠CBD)
=> ΔAPD ≅ ΔBQC
=> AP = BQ
=>∠APD = ∠CQB
=> ∠APQ = ∠CQP
=> AP ║ CQ
now join AC intersecting PQ at M
ΔAPM & ΔCQM
AP = CQ ( shown above)
∠APQ = ∠CQP ( shown above)
∠AMP = ∠CMQ (opposit angles)
=> ΔAPM ≅ ΔCQM
=> PM = QM
=> M is mid point of PQ
=> AC bisects PQ
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Answer:
We know that, diagonals of a parallelogram bisect each other OA = OC and OB = OD Since P and Q are point of intersection of BD BP = PQ = QD Now, OB = OD and BP = QD OB - BP = OD - QD OP = OQ Thus in quadrilateral APCQ, we have OA = OC and OP = OQ diagonals of quadrilateral APCQ bisect each other APCQ is a parallelogram Hence AP || CQ
