Diagonals AC and BD of trapezium ABCD with AB||DC intersect each other at O.
prove that ar(AOD) = ar (BOC)?
Given that ABCD is a trapezium with AB || DC and 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)
Please post single query in a thread
Given : ABCD is trapezium, AB parallel to DC, diagonals AC & BD intersect at O
To Prove : ar( AOD ) = ( BOC)
Proof : Since triangle ADC & BCD lies on same base DC and between same parallels AB & DC
ar ( ADC ) = ar ( BCD )
Subtracting ar ( DOC ) from both the sides
ar ( ADC ) - ar ( DOC ) = ar ( BCD ) - ar ( DOC )
ar ( AOD ) = ar ( BOC )