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.