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
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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.
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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