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.

Asked by Pragya Singh | 1 year ago |  57

##### Solution :-

Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,

$$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$.

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 ΔDAB, EO || AB

Therefore, By using Basic Proportionality Theorem

$$\dfrac{DE}{EA}=\dfrac{DO}{OB}$$ ……………………(i)

Also, given,

$$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$

⇒ $$\dfrac{AO}{CO}=\dfrac{BO}{DO}$$

⇒ $$\dfrac{CO}{AO}=\dfrac{DO}{BO}$$

$$\dfrac{DO}{OB}=\dfrac{CO}{AO}$$ …………………………..(ii)

From equation (i) and (ii), we get

$$\dfrac{DE}{EA}=\dfrac{CO}{AO}$$

Therefore, By using converse of Basic Proportionality Theorem,

EO || DC also EO || AB

⇒ AB || DC.

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

Answered by Abhisek | 1 year ago

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