Question

Abcd is a trapezium in which ab parallel to dc and its diaginals intersect each other at o. Prove that AO by BO equal to CO by DO

Answer 1

Hey there,

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)

Hope this helps!