The diagonals of a quadrilateral ABCD intersect each other at point O. Show that AO/BO = CO/DO.
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Answers
Answer: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.
Step-by-step explanation:
Step-by-step explanation:
Given:
The diagonals of a quadrilateral ABCD intersect each other at the point O such that
AO/BO=DO/CO
i.e.,
AO/CO= DO/BO
To Prove: ABCD is a trapezium
Construction:
Draw OE∥DC such that E lies on BC.
Proof:
In △BDC,
By Basic Proportionality Theorem,
DO/BO= BE/EC ............(1)
But,
AO/CO=DO/BO (Given) .........(2)
from. (1)&(2)
AO/CO= BE/EC
Hence, By Converse of Basic Proportionality Theorem,
OE || AB
since AB||CD||OE
AB||CD
hence ABCD is a trapezium