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If a hexagon ABCDEF circumscribe a circle, prove that `AB + CD + EF=BC+DE+FA` |
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Answer» We know that the tangents drawn from an external point to a circle are equal. `:." "AP=AV" "BP=BQ` `CR=QC" "DR=SD` `ET=ES" "TF=FU` Adding, we get ` AP+BP+CR+DR+ET+TF=AU+BQ+QC+SD+ES+FU` `implies" "AB+CD+EF=(AU+FU)+(BQ+QC)+(SD+ES)` `=AF+BC+DE` Hence Proved. |
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