Answer:
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 the basic proportionality theorem
\(\frac{AE}{ED} = \frac{AO}{OC}\) ………………………………(1)Now consider Δ ABD
where EO ll AB
According to the basic proportionality theorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii), we get,
AO/CO = BO/DO
⇒AO/BO = CO/DO
Hence, proved.
Hence the proof.
