the semi perimeter of triangle ABC=28 then AF+ bd +ce=
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Since length of tangents from an external point to a circle are equal
Therefore AF=AD
DB=BE
CF=CE
Since semiperimeter of ΔABC=28cm
⇒ perimeter of ΔABC=56cm
⇒AB+BC+AC=56
(AF+DB)+(DB+CE)+(AF+CE)=56
2(AF+BD+CE)=56
AF+BD+CE=28cm
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