to: The diagomts of a iquadrilateral ABCD inersecr oncly oler at the p

1.	to: The diagomts of a iquadrilateral ABCD inersecr oncly oler at the paix O such ileiAO coBO DO Show stui ABCDisattrape/ium
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.

to: The diagomts of a iquadrilateral ABCD inersecr oncly oler at the paix O such ileiAO coBO DO Show stui ABCDisattrape/ium

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.