Question
If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.
✓
Answer
Given: A circle inscribed in a hexagon ABCDEF.
Sides, AB, BC, CD, DE and DF touches the circle at P, Q, R, S, T and U respectively.
To prove: AB + CD + EF = BC + DE + FA
Proof: We know that tangents from an external point to a circle are equal.
Here, vertices of hexagon are outside the circle so
AP = AU
BP = BQ
CQ = CR
DR = DS
ES = ET
FT = FU
LHS = AB + CD + EF = (AP + PB) + (DR + CR) + (ET + TF)
By using above results, we have
LHS = AB + CD + EF = AU + BQ + DS + CQ + ES + FU
= AU + FU + BQ + CQ +DS + ES
= AF + BC + DE.
Hence, proved.
Sides, AB, BC, CD, DE and DF touches the circle at P, Q, R, S, T and U respectively.
To prove: AB + CD + EF = BC + DE + FA
Proof: We know that tangents from an external point to a circle are equal.
Here, vertices of hexagon are outside the circle so
AP = AU
BP = BQ
CQ = CR
DR = DS
ES = ET
FT = FU
LHS = AB + CD + EF = (AP + PB) + (DR + CR) + (ET + TF)
By using above results, we have
LHS = AB + CD + EF = AU + BQ + DS + CQ + ES + FU
= AU + FU + BQ + CQ +DS + ES
= AF + BC + DE.
Hence, proved.
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