Question

10. The diagonals of a quadrilateral ABCD intersect each other at the point o such that AO/BO = CO/DO. Show that ABCD is a trapezium.


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Answer 1

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

AO/BO = CO/DO.

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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

DE/EA = DO/OB ……………………(i)

Also, given,

AO/BO = CO/DO

⇒ AO/CO = BO/DO

⇒ CO/AO = DO/BO

⇒DO/OB = CO/AO …………………………..(ii)

From equation (i) and (ii), we get

DE/EA = 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.

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figure in attachment

Answer 2

Given:-

• The diagonals of a quadrilateral ABCD intersect each other at the point O such that,

$\sf{\bold{ \frac{AO}{BO} = \frac{CO}{DO} }}$

$\sf{\bold{i.e. \: \frac{AO}{CO} = \frac{BO}{DO} }}$

To Prove:-

• ABCD is a trapezium.

Construction:-

• Draw OEllDC such that E lies on BC.

Proof:-

In triangle BDC,

By basic proportionally theorem,

$\implies\sf{ \frac{BO}{OD} = \frac{BE}{EC}----(1)}$

$\implies\sf{But, \frac{AO}{CO} = \frac{BO}{DO}----(1)}$

From (1) and (2)

$\implies\sf{ \frac{AO}{CO} = \frac{BE}{EC}}$

Hence, By converse of basic proportanilly theorem, OEllAB.

Now, Since ABllOEllDC. So, ABllDC.

Hence, ABCD is a trapezium.