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Answer :
Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.
We have to prove, \(\frac{AO}{BO} = \frac{CO}{DO}\)
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In \(\triangle ADC\), we have OE || DC
Therefore, By using Basic Proportionality Theorem
\(\frac{AE}{ED} = \frac{AO}{CO}\) ……………..(i)
Now, In \(\triangle ABD\), OE || AB
Therefore, By using Basic Proportionality Theorem
\(\frac{DE}{EA} = \frac{DO}{BO}\)…………….(ii)
From equation (i) and (ii), we get,
\(\frac{AO}{CO} = \frac{BO}{DO}\)
=>\(\frac{AO}{BO} = \frac{CO}{DO}\)
Hence, proved.