Through A, B and C, lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a D ABC a shown in Fig.8.5. Show that BC = QR.



Given,


PQ||AB, PR||AC and RQ||BC.


In quadrilateral BCAR,


BR||CA and BC||RA


BCAR is a parallelogram


BC = AR …(i)


Now, in quadrilateral BCQA,


BC||AQ and AB||QC


BCQA is a parallelogram


BC = AQ …(ii)


Adding Eqn. (i) and (ii), we get


2BC = AR + AQ


2BC = RQ



Hence, proved.


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