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

1 Answer

Solution :-

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

\( \dfrac{AO}{BO}=\dfrac{CO}{DO}\).

Ncert solutions class 10 chapter 6-13

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


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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