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927 ABCDE Sa penagon. A line thnughHDACmees De prodaced at F Show |
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Answer» given: BF||AC(i)we know that if Triangles lie on the same base and between the same parallels then their areas are equal. Triangles ACB and ACF are on the same base AC and between the same parallels AC and BFTherefore, ar(ACB)=ar(ACF) --------(1)hence proved. (ii)adding ar AEDF to both sides of (1)ar(ACB)+ar(AEDF)=ar(ACF)+ar(AEDF)=>ar(ABCDE)=ar(AEDF)Hence proved. |
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