Q10... The person who will give me correct answer fast will be marked as brainliest....
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okay see
triangles AOD and AOB lie on equal bases DO and BO and between same parallels BD and A.
therefore ar(AOD) = ar(AOB) ------1
Similarly triangles COD and COB lie on equal bases DO and BO and between same parallels BD and C
therefore ar(COD) = ar(COB) ------2
Adding 1 and 2,
ar(AOD) + ar(COD) = ar(AOB) + ar(COB)
ar(ADC) = ar(ABC)
hence proved
Hope this solution helped!
triangles AOD and AOB lie on equal bases DO and BO and between same parallels BD and A.
therefore ar(AOD) = ar(AOB) ------1
Similarly triangles COD and COB lie on equal bases DO and BO and between same parallels BD and C
therefore ar(COD) = ar(COB) ------2
Adding 1 and 2,
ar(AOD) + ar(COD) = ar(AOB) + ar(COB)
ar(ADC) = ar(ABC)
hence proved
Hope this solution helped!
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