ABCD is a parallelogram. AP bisects ZA and CQ bisects ZC. P lies on CD and Q
lies on AB. Show that
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(6) AOCP is a parallelogram.
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Since the circum-centres of the △ABC and △ADC i.e.. Q and R are the points of concurrence of the perpendicular bisectors of their sides. If O is the mid-point of Diagonal AC, then
∴ Diagonals of quadrilateral AQCR bisect each other respectively at right-angle and MQ and NR are the perpendicular bisectors of BC and AD. Again, the circum-centre of a triangle is equidistant from its vertices.
∴AQ=CQ=BQ
and AR=CR=DR
Hence, the quadrilateral AQCR is always a rhombus.
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