1.

If a hexagon ABCDEF circumscribe a circle, prove that `AB + CD + EF=BC+DE+FA`

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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