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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1. 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 ☺️

   Reply

2. Using Theorem:-

   Two Triangles on the same base and between

   the same parallels are equal in area.

   ===========================================

   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)

   ==========================================================

   Hope this will help you…!!

   Aree thak mast ree xD

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