ABCD is a trapezium in which AB// DC and its diagonals intersect each other at the point o.show that AO/OB=OC/OD.
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Given: □ABCD is a trapezium where, AB ll CD
Diagonals AC and BD intersect at point O.
Construction: Draw a line EF passing through O and also parallel to AB.
Now, AB ll CD, since by construction, EF ll AB ⇒ EF ll CD
Consider the ΔADC,
EO ll DC
Thus, by Basic proportionality theorem, (AE / ED) = (AO / OC) .... (i)
Now, consider Δ ABD,
EO ll AB,
Thus, by Basic proportionality theorem, (AE / ED) = (BO / OD) .... (ii)
From (i) and (ii), we have, (AO / OC) = (BO / OD) (since L.H.S of i and ii are equal)
Hence we proved that, (AO / OC) = (BO / OD)
hopes this helps u !!!!!
Diagonals AC and BD intersect at point O.
Construction: Draw a line EF passing through O and also parallel to AB.
Now, AB ll CD, since by construction, EF ll AB ⇒ EF ll CD
Consider the ΔADC,
EO ll DC
Thus, by Basic proportionality theorem, (AE / ED) = (AO / OC) .... (i)
Now, consider Δ ABD,
EO ll AB,
Thus, by Basic proportionality theorem, (AE / ED) = (BO / OD) .... (ii)
From (i) and (ii), we have, (AO / OC) = (BO / OD) (since L.H.S of i and ii are equal)
Hence we proved that, (AO / OC) = (BO / OD)
hopes this helps u !!!!!
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