abcd is a trapezium in which ab parallel dc and its diagonals intersect each other at point o . show that ao/bo=co/do?? this is a textual exercize question can we do it using similarity (aa)???
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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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Answer:
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)
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