Question

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

Solution

## Given:The diagonals of a quadrilateral ABCD intersect each other at the point O such that $$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$i.e., $$\dfrac{AO}{CO}=\dfrac{BO}{DO}$$To Prove: $$ABCD$$ is a trapeziumConstruction:Draw $$OE\parallel DC$$ such that $$E$$ lies on $$BC.$$Proof:In $$\triangle BDC$$,By Basic Proportionality Theorem,$$\dfrac{BO}{OD}=\dfrac{BE}{EC}\,............(1)$$But, $$\dfrac{AO}{CO}=\dfrac{BO}{DO}$$ (Given) $$.........(2)$$$$\therefore$$ From $$(1)$$ and $$(2)$$$$\dfrac{AO}{CO}=\dfrac{BE}{EC}$$Hence, By Converse of Basic Proportionality Theorem,$$OE\parallel AB$$Now Since, $$AB\parallel OE\parallel DC$$$$\therefore$$ $$AB\parallel DC$$Hence, $$ABCD$$ is a trapezium.MathematicsRS AgarwalStandard X

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