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

Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,
$$\frac{AO}{BO} = \frac{CO}{DO}$$.

We have to prove here, ABCD is a trapezium

From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB

In $$\triangle DAB$$, EO || AB

Therefore, By using Basic Proportionality Theorem
$$\frac{DE}{EA} = \frac{DO}{OB}$$ ……………………(i)

Also, given,
$$\frac{AO}{BO} = \frac{CO}{DO}$$
=> $$\frac{AO}{CO} = \frac{BO}{DO}$$
=> $$\frac{CO}{AO} = \frac{DO}{BO}$$
=> $$\frac{DO}{OB} = \frac{CO}{AO}$$ …………………………..(ii)

From equation (i) and (ii), we get
$$\frac{DE}{EA} = \frac{CO}{AO}$$

Therefore, By using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
=> AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.