Question

The diagonals of a quadrilateral ABCD intersect each other at the point O such that `(AO)/(BO) = (CO)/(DO)`. Show that ABCD is a trapezium.

Answer 1

Let us consider the following figure for the given question.

Draw a line OE || AB

In ΔABD, OE || AB

By using the basic proportionality theorem, we obtain

`(AE)/(ED) = (BO)/(OD)`         ...(1)

However, it is given that

`(AO)/(OC) = (OB)/(OD)`                ...(2)

From equations (1) and (2,) we obtain

`(AE)/(ED) = (AO)/(OC)`

⇒ EO || DC             ...[By the converse of the basic proportionality theorem]

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.