In ؈ABCD, A-P-D, B-Q-C. If AB || PQ and PQ || DC, prove that AP*QC = PD*BQ.
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In □ABCD , A–P–D, B–Q–C. AB || PQ and PQ || DC.
we have to prove that AP. QC = PD.BQ
proof : AB || PQ and PQ || DC.
so, AB || PQ || DC.
so, AB, BC and DC are three parallel lines and AD and BC are their transversal.
∵ if three or more than three ∥ lines are intercepted by two transversal, the segments cut off on the transversal between the same parallel lines are proportional.
∴ AP/BQ = PD/QC
cross multiplying
∴ AP × QC = PD × BQ
hence proved
we have to prove that AP. QC = PD.BQ
proof : AB || PQ and PQ || DC.
so, AB || PQ || DC.
so, AB, BC and DC are three parallel lines and AD and BC are their transversal.
∵ if three or more than three ∥ lines are intercepted by two transversal, the segments cut off on the transversal between the same parallel lines are proportional.
∴ AP/BQ = PD/QC
cross multiplying
∴ AP × QC = PD × BQ
hence proved
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