In the figure, ABCD is a trapezium in which AB || DC, the diagonals AC & BD intersect at O. Prove that AO / OC = OB /DO.
Attachments:
Answers
Answered by
22
BASIC PROPORTIONALITY THEOREM (BPT) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. This is also known as Thales theorem.
SOLUTION:
GIVEN: ABCD is a trapezium in which AB || DC. Diagonals AC and BD intersect each other at O.
To Prove: AO/BO = CO/DO
Construction: Through O, draw EOF || AB || DC
Proof:
In ΔADC,
OE || DC (By Construction)
AE/ED = AO/CO …….……….…(i)
[By Basic Proportionality Theorem]
In ΔABD,
OE || AB (By Construction)
AE/ED = OB/OD………... …(ii)
[By Basic Proportionality Theorem]
From eq (i) and (ii)
AO/CO = OB/OD
Hence, proved.
HOPE THIS WILL HELP YOU...
SOLUTION:
GIVEN: ABCD is a trapezium in which AB || DC. Diagonals AC and BD intersect each other at O.
To Prove: AO/BO = CO/DO
Construction: Through O, draw EOF || AB || DC
Proof:
In ΔADC,
OE || DC (By Construction)
AE/ED = AO/CO …….……….…(i)
[By Basic Proportionality Theorem]
In ΔABD,
OE || AB (By Construction)
AE/ED = OB/OD………... …(ii)
[By Basic Proportionality Theorem]
From eq (i) and (ii)
AO/CO = OB/OD
Hence, proved.
HOPE THIS WILL HELP YOU...
Attachments:
Similar questions