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`Delta ABC and Delta DBC` lie on the same side of BC, show in the figure. From a point on `BC.PQ||AB and PR||BD` are drawn, meeting AC at Q, and CD at R respectively. Prove that `QR||AD`.

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In `Delta CAB, QP||AB rArr (CP)/(PB)=(CQ)/(QA)`
In `Delta CDB, RP||DB rArr (CP)/(PB)=(CR)/(RD)`
`:. (CQ)/(QA)=(CR)/(RD) rArr QR||AD ( "in" Delta CDA)`

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