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Through A, B and C, lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a \triangle ABC as shown in Figure. Show that BC=\frac{1}{2}QR

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Solution.
Given : In \triangle ABC, PQ\parallel ABand PR\parallel AC and PQ\parallel BC.
To Prove :
BC=\frac{1}{2}QR
Proof : In quadrilateral BCAR, BR\parallel CA and BC\parallel RA

Hence it is a parallelogram
BC = AR ……(1)
In quadrilateral BCQA,BC\parallel AQ and AB\parallel QC
Hence it is also a parallelogram
BC = AQ …..(2)|
Adding equations 1 and 2, we get
2BC = AR + AQ
2BC = RQ
BC=\frac{1}{2}QR
Hence proved

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