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In quad^ ABCD sideBC is parallel to sideAD.diagonal AC and diagonal BD intersect in point Q,if AO=1/3 AC then show that DQ=1/2BQ...plz guys help me..... |
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Answer» Answer: AQ= AC/3 AQ/QC = 1/2 AQ+QC = AC AC/3 + QC = AC QC = AC - AC/3 QC = 2/3 AC In triangle QAD and triangle QCB angle QAD =angle QCB (opposite alternate angle) angle QDA = angle QBC angle AQD = angle CQB ∆ QAD ~ ∆QCD AQ/QD = CQ/QB AQ/CQ = QD/QB QD/QB = 1/2 QD = QB/2 QD = 1/2 BQ hence PROVED Explanation: hope it helps yrr |
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