Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at O. Prove that diagonals of a rectangle is equal
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Using Theorem:-
Two Triangles on the same base and between
the same parallels are equal in area.
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Given:-
Diagonals AC and BD of a trapezium ABCD
with AB || DC intersect each other at O.
To Prove:-
ar (AOD) = ar (BOC).
Proof:-
Here, △DAC and △DBC lie on the same base DC and between thesame parallels AB and CD.
∴ ar(△DAC) = ar(△DBC)
ar(△DAC) − ar(△DOC) = ar(△DBC) − ar(△DOC)
[On subtracting ar(△DOC) from both sides]
ar(△AOD) = ar(△BOC)
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Hope this will help you...!!
Answered by
0
GIVEN THAT
- ABCD is a trapezium with AB || DC
- Diagonal AC and BD intersect each other at O.
TO PROVE
- Area (AOD) = Area (BOC)
PROOF
- ΔADC and ΔBDC are on the same base DC and between same parallel AB and DC.
- ∴Area (ΔADC) = Area (ΔBDC) [triangles on the same base and between same parallel are equal in area]
- Subtract Area (ΔDOC) from both side
- Area (ΔADC) – Area (ΔDOC) = Area (ΔBDC) – Area (ΔDOC)
- Area (ΔAOD) = Area (ΔBOC)
- Hence proved.
HOPE this helps you ☺️
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