Question

in a trapezium ABCD, AB||CD, AC and BD diagonds are intersect at o. prove that AO. OD=BO.OC​

Answer 1

Given parameters

ABCD is a trapezium where AB || DC and diagonals AC and BD intersect at O.

To prove

$\frac{AO}{BO} = \frac{CO}{DO}$

Construction

Draw a line EF passing through O and also parallel to AB

Now, AB ll CD

By construction EF ll AB

∴ EF ll CD

Consider the ΔADC,

Where EO ll AB

According to basic proportionality theorem

$\frac{AE}{ED} = \frac{AO}{OC} …………………(1)$

Now consider Δ ABD

where EO ll AB

According to basic proportionality theorem

$\frac{AE}{ED} = \frac{BO}{OD} ………………..(2)$

From equation (1) and (2) we have

$\frac{AO}{OC} = \frac{BO}{OD}$

$⇒ \frac{AO}{BO} = \frac{OC}{OD}$

Hence the proof.