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03/06/2021 12:06 pm
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ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
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03/06/2021 12:09 pm
Given, ABCD is a trapezium where AB || DC and diagonals 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 Proportionality Theorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii), we get,
AO/CO = BO/DO
⇒ AO/BO = CO/DO
Hence, proved.