# The diagonals of a quadrilateral ABCD intersect each other at the point O such that $$AO\over BO$$ = $$CO\over DO$$. Show that ABCD is a trapezium.

## Solution :

Given : A quadrilateral ABCD, in which the diagonals AC and BD intersect each other at O such that $$AO\over BO$$ = $$CO\over DO$$.

To  Prove : ABCD is a trapezium.

Construction : Draw EO || BA, meeting AD in E.

Proof : In triangle ABD, EO || BA

By basic proportionality theorem,

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

But Given that, $$AO\over BO$$ = $$CO\over DO$$ $$\implies$$  $$DO\over BO$$ = $$CO\over AO$$               …………..(2)

From (1) and (2), we get

$$DE\over EA$$ = $$CO\over OA$$

By converse of basic proportionality theorem,

$$\implies$$ EO || DC

But by construction, EO || BA

$$\therefore$$  DC || BA

Hence, ABCD is a trapezium.