# ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $$AO\over BO$$ = $$CO\over DO$$.

## Solution :

Given : A trapezium ABCD, in which the diagonals AC and BD intersect each other at O.

To  Prove : $$AO\over BO$$ = $$CO\over DO$$

Construction : Draw EF || BA || CD, meeting AD in E.

Proof : In triangle ABD, EF || AB

By basic proportionality theorem,

$$DO\over OB$$ = $$DE\over AE$$                    …………(1)

In triangle CDA, EO || DC,

By basic proportionality theorem,

$$CO\over OA$$ = $$DE\over AE$$                    …………..(2)

From (1) and (2), we get

$$DO\over OB$$ = $$CO\over OA$$

$$\implies$$    $$AO\over BO$$ = $$CO\over DO$$