Question -
Answer -
Given, ABCD is a trapezium where AB || DC anddiagonals AC and BD intersect each other at O.
We have to prove, AO/BO = CO/DO
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔADC, we have OE || DC
Therefore, By using Basic ProportionalityTheorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE || AB
Therefore, By using Basic ProportionalityTheorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii),we get,
AO/CO = BO/DO
⇒AO/BO = CO/DO
Hence, proved.