Question

ABCD is a trapezium in which AB parallel DC and its diagnols intersect each other at the point O. Show that AO/BO=CO/DO

Answer 1

Given: ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.
To Prove: AO/BO = CO/DO
Construction: Through O, draw EO || DC || AB
Proof: In ΔADC, we have
OE || DC (By Construction)
∴ AE/ED = AO/CO  ...(i) [By using Basic Proportionality Theorem]
In ΔABD, we have
OE || AB (By Construction)
∴ DE/EA = DO/BO ...(ii) [By using Basic Proportionality Theorem]
From equation (i) and (ii), we get
AO/CO = BO/DO
⇒  AO/BO = CO/DO