ABCD is a trapezium whose diagonal intersect at zero then prove that O

1.	ABCD is a trapezium whose diagonal intersect at zero then prove that OA/OB=OC/OD
Answer» Given: A trapezium ABCD, In which AB\xa0{tex}\\parallel{/tex} CD and its diagonals AC and BD intersect at O.To Prove: {tex}\\frac{{AO}}{{OC}} = \\frac{{BO}}{{OD}}{/tex}Construction: Through O draw OE\|\|ABproof: In {tex}\\triangle {/tex}ADC, OE {tex}\\parallel{/tex} DC,Hence {tex}\\frac{{AE}}{{ED}} = \\frac{{AO}}{{OC}}{/tex} .....(i).....[By BPT]Again in {tex}\\triangle {/tex}ABD, we have OE\xa0{tex}\\parallel{/tex} AB,hence {tex}\\frac{{DE}}{{EA}} = \\frac{{DO}}{{OB}}{/tex} ........[By BPT]{tex}\\Rightarrow {/tex}{tex}\\frac{{EA}}{{ED}} = \\frac{{OB}}{{OD}}{/tex} .....(ii)......[Using invertendo]{tex}\\therefore {/tex} From (i) and (ii), we.{tex}\\frac{{OB}}{{OD}} = \\frac{{AO}}{{OC}}{/tex}H once proved

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