Question

the diagonals of a quadrilateral ABCD intersect each other at the point O such that AO divided by BO equals to CO devided By DO show that ABCD is a trapezium

Answer 1

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.

Answer 2

I hope it will help you dear...