ABCD is a trapezium in which …
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Sia ? 6 years, 6 months ago
Given: A trapezium ABCD, In which AB {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||AB
proof: 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 {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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