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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 trapezium
Construction:
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.

493944_465426_ans_cabefe8118804d8488defd358ac22780.png

Mathematics
RS Agarwal
Standard X

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