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# ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $$\frac{AO}{BO} = \frac{CO}{DO}$$.

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.