Question

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

Answer 1

Answer:

Step-by-step explanation:

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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.