Question

The diagonals of a Trapezium ABCD in which DC is parallel to AB and its diagonal intersect each other at O then show that AO×OD = OB×OC

Answer 1

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