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ABCD is a parallelogram and O is a point in its interior. Prove that (i) `ar(triangleAOB)+ar(triangleCOD)` `=(1)/(2)ar("||gm ABCD").` (ii) `ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)`. |
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Answer» GIVEN A||gm ABCD and O is a point in its interior. TO PROVE (i) `ar(triangleAOB)+ar(triangleCOD)=(1)/(2)ar("||gm ABCD"),` (ii) `ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)`. CONSTRUCTION Draw EOF||AB||DC and GOH||AD||BC. PROOF `triangle`AOB and ||gm EABF being on the same base AB and between the same parallels AB and EF, we have `ar(triangleAOB)=(1)/(2)ar("||gm EABF")." "...(i)` Similarly, `ar(triangle COD)=(1)/(2)ar("||gm EFCD")," "...(ii)` `ar(triangleAOD)=(1)/(2)ar("||gm AHGD")" "...(iii)` and `ar(triangleBOC)=(1)/(2)ar("||gm BCGH")" "...(iv)` Adding (i) and (ii), we get `ar(triangleAOB)+ar(triangleCOD)=(1)/(2)ar("||gm EABF")+(1)/(2)ar("||gm EFCD")` `=(1)/(2)ar("||gm ABCD")." "...(v)` Adding (iii) and (iv), we get `aar(triangleAOD)+ar(triangleBOC)=(1)/(2)ar("||gm AHGD")+(1)/(2)ar("||gm BCGH")` `=(1)/(2)ar("||gm ABCD")." "...(vi)` `therefore ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)` [from (v) and (vi)]. |
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