Solve this:

Q.1  If a hexagon ABCDEF circumscribe a circle, prove that
             AB + CD + EF = BC + DE + FA

Sol.  Given A hexagon ABCDEF circumscribe a circle. 
 

here ABCDEF is a hexagon which circumscribe a circle.

since the tangents drawn from a point to the circle are equal in length.

therefore AM=AR,

BM=BN,

CN=CO,

DO=DP,

EP=EQ,

FQ=FR.

now from these equations we have,

AM+BM=AR+BN⇒AB=AR+BN

CO+DO=CN+DP⇒CD=CN+DP

EQ+FQ=EP+FR⇒EF=EP+FR

now on addition these three equation gives

AB+CD+EF=AR+(BN+CN)+(DP+EP)+FR=BC+DE+FA

hence proved.

Regards