o is the point of intersection of diagonals ac and bd of trapezium abcd with ab parallel dc . through o , pq is drawn parallel to ab prove that po=oq
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O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q, prove that PO = QO.
In ∆ABD and ∆POD, PO || AB [∵ PQ || AB] ∠D = ∠D [common angle] ∠ABD = ∠POD [corresponding angles] ∴ ∆ABD ∼ ∆POD [by AAA similarity criterion] Then, OP/AB = PD/AD …(i) [by basic proportionality theorem] In ∆ABC and ∆OQC, OQ || AB [∵ OQ || AB] ∠C = ∠C [common angle] ∠BAC = ∠QOC [corresponding angle] ∴ ∆ABC ∼ ∆OQC [by AAA similarity criterion] Then, OQ/AB = QC/BC …(ii) [by basic proportionality theorem] Now, in ∆ADC, OP || DC ∴ AP/PD = 0A/0C [by basic proportionality theorem] …(iii) In ∆ABC, OQ || AB ∴ BQ/QC = OA/OC [by basic proportionality theorem] …(iv) From Equation (iii) and (iv), AP/PD = BQ/QC Adding 1 on both sides, we get, = AP/PD + 1 = BQ/QC + 1 = ((AP + PD))/PD = (BQ + QC)/QC = AD/PD = BC/QC = PD/AD = QC/BC = OP/AB = OQ/BC [from Equation (i) and (ii)] ⇒ OP/AB = OQ/AB [from Equation (iii)] ⇒ OP = OQ Hence proved.
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