010. The diagonals of a quadrilateral ABCD intersect each other at the

1.	010. The diagonals of a quadrilateral ABCD intersect each other at the pointăsuch thatAO COBO DCShow that ABCD is a trapezium.
Answer» Given: Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that AO/BO = CO/DO. To Prove: ABCD is a trapezium Construction: Through O, draw line EO, where EO \|\| AB, which meets AD at E. Proof: In ΔDAB, we have EO \|\| AB ∴ DE/EA = DO/OB ...(i)[By using Basic Proportionality Theorem] Also, AO/BO = CO/DO (Given) ⇒ AO/CO = BO/DO ⇒ CO/AO = BO/DO ⇒ 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. Like my answer if you find it useful!

010. The diagonals of a quadrilateral ABCD intersect each other at the pointăsuch thatAO COBO DCShow that ABCD is a trapezium.

Answer»

Given: Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that AO/BO = CO/DO.

To Prove: ABCD is a trapezium

Construction: Through O, draw line EO, where EO || AB, which meets AD at E.

Proof: In ΔDAB, we have

EO || AB

∴ DE/EA = DO/OB ...(i)[By using Basic Proportionality Theorem]

Also, AO/BO = CO/DO (Given)

⇒ AO/CO = BO/DO

⇒ CO/AO = BO/DO

⇒ 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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