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Here DCIIAB
Therefore, ar (ADB) = ar(ACB)
Since these triangles are on same base AB and lie between same parallels DC and AB… Now,
ar (ADB)-ar(AOB)=ar(ACB)-ar(AOB)
if equals are subtracted from equals then the wholes are equals…
ar(AOD)=ar(BOC)
Thus proved…
Thank You…..
hope it helps....
Therefore, ar (ADB) = ar(ACB)
Since these triangles are on same base AB and lie between same parallels DC and AB… Now,
ar (ADB)-ar(AOB)=ar(ACB)-ar(AOB)
if equals are subtracted from equals then the wholes are equals…
ar(AOD)=ar(BOC)
Thus proved…
Thank You…..
hope it helps....
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